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Abstract
In this article we classify all L9-free finite groups.
Introduction
There are some algebraic laws that hold in a lattice L if and only if L does not have a sublattice of a specific shape. For example, a lattice is modular if and only if it does not have a sublattice isomorphic to the so-called pentagon L5.
If L is a lattice, then we call a group L-free if and only if its subgroup lattice does not contain a lattice isomorphic to L. For example, the finite L5-free groups are exactly the modular groups, and these have been classified by Iwasawa in 1941, see [Citation4]. The subgroup lattice of the dihedral group of order 8, often denoted by L10, and some of its sublattices are of particular interest. One reason is that, if p is a prime number, then a finite p-group is L5-free if and only if it is L10-free.
There are several sublattices of L10 containing L5:
In 1999 Baginski and Sakowicz [Citation2] studied finite groups that are L6-free and L7-free at the same time, and later Schmidt [Citation8] classified the finite groups that are L6- or L7-free. Together with Andreeva and the first author he also characterized, in [Citation1], all finite groups that are L8-free or M8-free. Finally, the finite M9-free groups have been classified by Pölzing and the second author in [Citation6]. Furthermore, there is a general discussion of L10-free groups by Schmidt, which can be found in [Citation9] and [Citation10].
In this paper we investigate finite L9-free groups. Since L9 is a sublattice of L10, the groups that we consider are L10-free and therefore we can use Corollary C in [Citation9] as a starting point for our analysis: Every finite L10-free group G has normal Hall subgroups such that
is a 2-group and
is meta-cyclic.
Our strategy is to choose maximal with respect to the above constraints, and then we show that N has a complement K that is a direct and coprime product of groups of the following structure: cyclic groups, groups isomorphic to Q8 or semi-direct products
, where Q has prime order and R is cyclic of prime-power order such that
. Furthermore,
is a 2-group and
is cyclic or elementary abelian of order 4 and every Sylow subgroup of
is elementary abelian or isomorphic to Q8. If the action of K on N satisfies some more conditions, then we say that NK is in class
.
The aim of our article is to prove the following theorem:
Main Theorem. A finite group is in class if and only if it is L9-free.
Notation and preliminary results
In this article we mostly follow the notation from Schmidt’s book [Citation7] and from [Citation5]. All groups considered are finite and G always denotes a finite group, moreover p and q always denote prime numbers. We quickly recall some standard concepts:
L(G) denotes the subgroup lattice of G, consisting of the set of subgroups of G with inclusion as the partial ordering. The infimum of two elements is
(their intersection) and the supremum is
(the subgroup generated by A and B).
If L is any lattice, then G is said to be L-free if and only if L(G) does not have any sub-lattice that is isomorphic to L.
A lattice L is said to be modular if and only if for all such that
, the following (also called the modular law) is true:
We say that a group G is modular if and only if L(G) is modular.
The modular law is similar to Dedekind’s law (see 1.1.11 of [Citation5]). For all such that
it says that
We will use Dedekind’s law frequently throughout this article without giving an explicit reference each time.
If , then we say that an element
induces power automorphisms on N if and only if
for all subgroups U of N. Furthermore
is a subgroup of G because
.
Lemma 1.1.
Let K be a finite group that acts coprimely on the p-group P. Then . If P is abelian, then this product is direct. If
, then
. Furthermore
and for all K-invariant normal subgroups N of P we have that
.
Proof.
These statements are a collection from 8.2.2, 8.2.7, 8.2.9, and 8.4.2 of [Citation5]. □
Lemma 1.2.
Let and suppose that G = PK, where P is a normal Sylow p-subgroup of G,
is a
-group and
. Suppose further that
and that K acts irreducibly on
. If
, then
.
Proof.
Let and
. Then we first remark that
. Lemma 1.1 shows that
and hence we have elements
and
such that g = ch. We note that
and therefore
is a normal subgroup of G, moreover P0 is a p-group and hence
is elementary abelian. Let
denote the natural homomorphism. As
is abelian and
acts coprimely on it, we see that
, again by Lemma 1.1. Therefore
. We recall that
and thus
, and then by hypothesis
and in particular
. It follows that
because
, see above. We conclude that
. By hypothesis K acts irreducibly on
, hence
does as well and we see that
. The main property of the Frattini subgroup (see for example 5.2.3 in [Citation5]) finally gives that
. □
Lemma 1.3.
Let Q be a 2-group that is elementary abelian, cyclic or isomorphic to Q8. Then Q does not admit power automorphisms of odd order.
Proof.
If Q is abelian, then the assertion follows from 2.2.5 of [Citation5]. If , then
, and any automorphism of order 3 interchanges the maximal subgroups of Q. □
Lemma 1.4.
Suppose that G = NK, where N is a normal Hall subgroup of G and K is a complement. Let and
. Then the following hold:
If
and
are subgroups of G, then
.
If
and
, then
.
If K is abelian and acts irreducibly on the abelian group
or if N is abelian and K induces power automorphisms on it, then
or
.
If
, then
.
Proof.
Suppose that and
are subgroups of G.
For (a) we do the following calculation:
Then (a) yields that . Therefore, if
and if
and
are such that
, then the fact that
implies that a = b and that
. We deduce that
and that
. Hence
. This is (b).
If K is abelian, then and
is K-invariant. Thus, if K acts irreducibly on
and
, then
, and the fact that
implies that
. If K induces power automorphisms on the abelian group N, then 1.5.4 of [Citation7] implies that these are universal. If x = 1, then
, and otherwise
and every element of K that centralizes x also centralizes N. Altogether (c) holds.
Suppose finally that . Then, for all
, we have that
. It follows that
.
On the other hand for all
and therefore
.
This implies that . Since
, we deduce that
, which is (d). □
2 Battens and batten groups
Definition 2.1.
We say that G is a batten if and only if G is a cyclic p-group, or isomorphic to Q8, or G = QR, where Q is a normal subgroup of prime order and R is a cyclic p-group of order coprime to
and such that
.
We say that G is a batten group if and only if G is a direct product of battens of pairwise coprime order.
If G is a batten group, then we say that
is a batten of G if and only if B is a batten that is one of the direct factors of G.
Warning: It is possible for a subgroup of a batten group G to be a batten, abstractly, but not to be a batten of G. This can happen when it is a p-subgroup for some prime p of a batten as in the third case of Definition 2.1.
Example 2.2.
Suppose that
is a group of order 19 and that
is a subgroup of
of order 27. Further suppose that
. Then
is a non-nilpotent batten. For this we calculate that
. Then the fact that
implies that
. We note that B is a batten group and that Q is a subgroup of B that is a batten, but not a batten of B because
.
Let B = QR be as in (a), let
and let S be a cyclic group of order 625. Then
is a batten group.
Remark 2.3.
Let G be a batten group and let B be a batten of G such that .
B is not nilpotent, but B has a unique normal Sylow subgroup.
A Sylow subgroup Q of B is cyclic, and therefore Q is batten. But Q is not a direct factor of G and hence Q is not a batten of G.
Definition 2.4.
Suppose that G is a non-nilpotent batten. Then there is a unique prime such that G has a normal Sylow q-subgroup Q, and Q is cyclic of order q. In this case we set
.
From the definition we can immediately see that is a characteristic subgroup of a non-nilpotent batten G and that it has prime order.
Lemma 2.5.
Suppose that G is a non-nilpotent batten, that and that
has order at least r 2. Then
.
Proof.
Since , we see that
. Then Definition 2.1 implies that R is cyclic and that there is a prime
such that
.
Now and
. We recall that R is cyclic, and then this implies that
. Since G is not nilpotent, we see that
is not contained in Z(G). Thus Z(G) is an r-group, because
has prime order. In addition
, because G is not nilpotent. Since
is a maximal subgroup of the cyclic group R, it follows that
. Moreover we have that
, whence
. This proves all statements. □
Lemma 2.6.
If G is a batten group and is a Sylow p-subgroup of G for some prime p, then G has a subgroup of order p. In addition,
or there is some non-nilpotent batten B of G such that
.
Proof.
Since , it follows that P is cyclic or isomorphic to Q8. Therefore
has order p.
If P is a batten, then . In particular
is the unique subgroup of G of its order. Otherwise there is a non-nilpotent batten B of G such that
. If P has order p, then
is a normal subgroup of B and so of G. Again
is the unique subgroup of K of order q. Otherwise we have that
by Lemma 2.5. In particular
is the unique subgroup of K of its order. □
Lemma 2.7.
Suppose that H is a batten and that . Then U is a cyclic batten group. Furthermore, all subgroups of batten groups are batten groups.
Proof.
Assume for a contradiction that U is not a cyclic batten group. Then U is not a cyclic batten, and therefore H is neither cyclic of prime power order nor isomorphic to Q8. Thus H is not nilpotent, in particular and all Sylow subgroups of H are cyclic. This implies that U is not a p-group. Let
, let
and
be such that H = QR and
. Now
as well and therefore
. Then Dedekind’s law gives that
is a proper subgroup of
, and it follows that
is a proper subgroup of R. In particular, since R is cyclic, we have that
. Then Lemma 2.5 gives that
. Altogether
. But then U is a direct product of cyclic groups of prime power order, i.e. a cyclic batten group, and this is a contradiction.
Next suppose that G is a batten group and that U is a subgroup of G. Then U is a direct product of subgroups of the battens of G whose orders are pairwise coprime. Consequently U is a batten group as well, by the arguments above. □
We remark that sections of battens, or batten groups, are not necessarily batten groups. For example, is not a batten group.
Lemma 2.8.
Suppose that K is a batten group and that is a q-group.
If Q is not normal in K, then there is a non-nilpotent batten B of K such that and
.
If , then
or this index is a prime number.
Proof.
Since K is a batten group, there is a batten B of K such that . Moreover there is a subgroup L of K such that
. Then
(*).
We first suppose that Q is not a normal subgroup of K. Since K is a direct product of battens, it follows that Q is not normal in B. Thus B is neither abelian nor hamiltonian (otherwise all subgroups of B would be normal), and it follows that B is not nilpotent. Now Q is a proper subgroup of B because . We conclude from Lemma 2.5 that neither
nor
, whence
and therefore
. Consequently (*) and Dedekind’s modular law yield that
.
Suppose now that . Using (*) we see that
. Hence we may suppose that B is not abelian. If B is not nilpotent, then Z(B) and a Sylow q-subgroup of B centralize Q. Thus Lemma 2.5 yields that
equals the prime in
. Let
and suppose that
. Then Q has order 4 or 8. In the first case
and in the second case
, which completes the proof. □
3 L10 and its sublattices
Throughout this article we will use the notation from the next definition whenever we refer to L10 and its sublattices:
Definition 3.1.
The lattice L10 is defined to be isomorphic to , with notation as indicated in the picture.
Now we define
,
,
,
,
,
,
,
Definition 3.2.
Let L be a lattice. An equivalence relation on L is called a congruence relation if and only if, for all
such that
and
, we have that
and
.
Lemma 3.3.
Let be a congruence relation on
as in Definition 3.1 and suppose that
is not equality. Then
.
Proof.
Let be a congruence relation on L9 and suppose that
.
If , then
and therefore
.
First we assume that is such that
. Then there is some
such that
, and then
. We choose
such that
. Then
and therefore, if
, then
. Now it follows that
, which gives a contradiction.
We have seen that E is not congruent to any of the elements . Next we assume that
is such that
and we choose
such that
. Then
, which gives another contradiction. We conclude that {E} and {F} are singleton classes with respect to
.
If there are such that
and
or
, then this implies that
or
. As this is impossible, we conclude that such elements
do not exist.
In particular are pairwise non-congruent and
are also pairwise non-congruent.
We assume that . Then we choose
such that
, and this gives the contradiction
. Hence {D} is a singleton as well.
Finally, we assume that there are and
such that
. We have seen that
and then it follows that
by the structure of L9. Now
, which is impossible. In conclusion, for all
, we have that
if and only if X = Y. This means that
is equality. □
In the following lemma we argue similarly to Lemma 2.2 in [Citation8].
Lemma 3.4.
Suppose that and that
are normal subgroups of G of pair-wise coprime order such that
. Then G is L9-free if and only if, for every
, the group Gi is L9-free.
Proof.
Since subgroups of L9-free groups are L9-free we just need to verify the “if” part.
Suppose that G1,…,Gn are L9-free. Then Lemma 1.6.4 of [Citation7] implies that . By induction we may suppose that n = 2. Assume that
is a sublattice of
that is isomorphic to L9 as in Definition 3.1. Then the projections
and
of L into
and
, respectively, are not injective, because
and
are L9-free. Let
and define, for all
:
Then is a congruence relation on L, because
is a lattice homomorphism, but it is not equality because
is not injective. Then Lemma 3.3 implies that
and
, and hence D = E. This is a contradiction. □
Lemma 3.5.
as in Definition 3.1 is completely characterized by the following:
L9(i)
.
L9(ii)
and
.
L9(iii)
and
.
L9(iv)
.
L9(v)
and
.
Proof.
We first remark that L9 satisfies the relations given in L9 (i) – L9 (v).
Suppose conversely that a lattice satisfies the relations given in L9 (i) – L9 (v). Then we see that
and
as well as
and
. If these are the unique inclusions and
, then
.
For all such that
we have that
and
. Thus L9 (ii) shows that S, T, and D are pair-wise not subgroups of each other.
Using L9 (iii) we obtain that and
, and L9 (iv) gives that also S, T and U are pair-wise not subgroups of each other. In addition, by L9 (v), we have that A, B and C are pair-wise not subgroups of each other. Together with the fact that
and
, this implies that
and
. Moreover we have that
and
and
and
and
. Together with L9 (ii) and L9 (iii), this information yields that
and
as well as
.
We conclude that there is a lattice homomorphism from L9 to L. Hence we obtain a congruence relation
on L9 by defining that
if and only if
, for all
.
If is not injective, then Lemma 3.3 implies that E = D. This contradicts L9 (i). Consequently
is injective and
. □
The next lemma gives an example of a group that is not L9-free.
Lemma 3.6.
D12 is not L9-free.
Proof.
Let G be isomorphic to D12 and let be such that o(a) = 6, o(b) = 2 and
. Then we find a sublattice in L(G) isomorphic to L9 by checking the equations from Lemma 3.5.
We let and we define
and
.
L9(i)We see that and hence
.
L9(ii)We notice that is isomorphic to
with cyclic normal subgroup
of order 3 and distinct subgroups
of order 2. Then
and
.
L9(iii)The subgroup C is also isomorphic to , the subgroup
has order 2, and moreover
and
.
L9(iv)We first see that and then
.
L9(v) is a cyclic normal subgroup of G of order 6, and the subgroups A and C also have order 6. These three subgroups are maximal in G. Hence
. Assume for a contradiction that A = C. Then
, which is false. Since
is the unique subgroup of order 3 of G, we conclude that
.
Altogether it follows, with Lemma 3.5, that L is isomorphic to L9, and then G is not L9-free. □
The next lemma shows how we can construct an entire class of groups that are not L9-free.
Lemma 3.7.
Suppose that and that G = PQ, where P is an elementary abelian normal Sylow p-subgroup of G and Q is a cyclic Sylow q-subgroup of G. Suppose that Q acts irreducibly on
and that
. Then G is not L9-free.
Proof.
Since Q is abelian, we see that . We claim that
is not L9-free. Therefore we may suppose that E = 1.
Our hypotheses imply that is not centralized by Q, and in particular
. Moreover
by hypothesis. Since P is elementary abelian, Lemma 1.1 gives that
.
We let and
be subgroups of minimal order such that
and
and we set
. If p is odd, then A has order p2, and if p = 2, then
. In the first case A has
subgroups isomorphic to V. In the second case A has
subgroups isomorphic to V, where
of these subgroups intersect V non-trivially and 19 of them intersect D non-trivially. In both cases, we find subgroups T and S of A isomorphic to V such that
and
.
We recall that A is elementary abelian, and then it follows that L9 (i) and L9 (ii) hold and that (*).
We further set . Then
and
, which implies L9 (iii).
If , then
and then the irreducible action of Q on
and Lemma 1.2 yield that
. Using (*) it follows that
. Combining all this information gives that
. Now if we set
, then we have L9 (iv).
To prove our claim, it remains to show that property L9 (v) of Lemma 3.5 is satisfied.
We set for some
.
If , then
and hence
. We further have that
, by Lemma 1.4 (b) and (c), because Q acts irreducibly on
.
In addition, the irreducible action of Qx on and Lemma 1.2 yield that
. It follows that
. Finally we deduce from Part (d) of Lemma 1.4 that
□
4 Group orders with few prime divisors
Much of our analysis will focus on non-nilpotent groups with a small number of primes dividing their orders. The next lemma sheds some light on why this situation naturally occurs.
Lemma 4.1.
Suppose that G is L9-free. Then G possesses a normal Sylow subgroup.
Proof.
Assume that this is false. Since G is L9-free and hence L10-free, [Citation9, Corollary C] is applicable. Then G is metacyclic because it does not have any normal Sylow subgroup, and it follows that G is supersoluble. Then Satz VI.9.1(c) in [Citation3] gives a contradiction. □
Lemma 4.2.
Suppose that G is a p-group. Then the following statements are equivalent:
L(G) is modular.
G is L5-free.
G is L9-free.
G is L10-free.
Proof.
This lemma follows from Theorem 2.1.2 in [Citation7] and Lemma 2.1 in [Citation9], since L9 is a sublattice of L10 containing L5. □
Lemma 4.3.
Suppose that and that G is an L9-free {p, q}-group. Let P be a normal Sylow p-subgroup of G and let
. If G is not nilpotent, then Q is cyclic or
.
Proof.
First we note that all subgroups and sections of G are L9-free and that G is L10-free.
Let G be non-nilpotent and assume for a contradiction that Q is neither cyclic nor isomorphic to Q8. Given that G is L10-free, we may apply Theorem B of [Citation9] and we see that neither (a), (b) nor (c) hold. Therefore p = 3 and q = 2. Now there are such that
is not cyclic and b is an involution. If, for all choices of b, we have that
, then
acts element-wise fixed-point-freely on
, contradicting 8.3.4 (b) of [Citation5]. Therefore we may choose b such that a does not centralize
, and we also choose a of minimal order under these constraints. Then a2 centralizes P and a inverts an element
by a result of Baer (e.g. 6.7.7 of [Citation5]). If follows that a inverts
and we may suppose that x has order 3. Now
is isomorphic to D12, contrary to Lemma 3.6. □
Lemma 4.4.
Suppose that and that G is an L9-free {p, q}-group. Furthermore, let P be a normal Sylow p-subgroup of G and let
be cyclic such that
is elementary abelian.
Then every subgroup of Q acts irreducibly or by inducing (possibly trivial) power automorphism on . Moreover,
is a cyclic 2-group and P is abelian.
Proof.
First we note that all subgroups and sections of G are L9-free and that all subgroups and sections of are elementary abelian, by hypothesis. In addition Lemma 2.2 of [Citation9] yields that
, as G is also L10-free. In particular P is abelian if
is.
Assume that the lemma is false and let G be a minimal counterexample.
Since is elementary abelian, we introduce the following notation with Maschke’s theorem:
Let and let
be Q-invariant and such that
and that Q acts irreducibly on
, respectively. Lemmas 2.3.5 of [Citation7] and 4.2 yield that
is elementary abelian. Now there are
and cyclic subgroups
of
such that
.
We set and
.
Then for every the group
is elementary abelian. Moreover Hi is a proper subgroup of G and then the minimal choice of G implies that every subgroup of Q either induces (possibly trivial) power automorphisms on
or acts irreducibly on it.
(1) and
.
Proof. We assume for a contradiction that .
Then Q does not act irreducibly on both and
, and it follows that Q induces (possibly trivial) power automorphisms on
.
We suppose first that . Then
for both
. Therefore Lemma 1.5.4 of [Citation7], together with the fact that
, provides some
such that
for every
. But this means that Q, and hence every subgroup of Q, induces (possibly trivial) power automorphism on
in this case. Thus G is not a counterexample, which is a contradiction.
We conclude that and now there is some
such that
and
. Then Hi satisfies the hypotheses of our lemma and it follows that
is an non-trivial 2-group. In particular p = 2. But then Lemma 1.3 provides the contradiction that
.
For the proof of (1), we assume for a further contradiction that . Then Q acts irreducibly on the elementary abelian group
and Lemma 3.7, applied to
, gives that
. In particular we have that p = 2. Thus the minimal choice of G and Lemma 1.3 yield, for every proper subgroup U of Q, that U centralizes P or acts irreducibly on
.
Since G is a counterexample, it follows that is not cyclic. But
and therefore
is a generalized quaternion group. It follows that
by Lemma 4.2. In this case
and G/Z satisfies the hypotheses of our lemma, but not the conclusion. Thus G is not a minimal counterexample, contrary to our choice. □
(2) Q acts irreducibly on .
Proof. Assume for a contradiction that n = 2. By hypothesis Q is cyclic, and then we may suppose that . If
, then Lemma 2.8 of [Citation9] implies that Q induces power automorphisms on P. Thus G is not a counterexample, which is a contradiction.
Therefore and
. The minimal choice of G yields that Q0 acts irreducibly or by inducing power automorphisms on
and that
is a cyclic 2-group. Now we notice that
, but
, whence we deduce a contradiction from 2.2.5 of [Citation5]. □
Since G = PQ is a counterexample to the lemma, Q has a proper subgroup U that does not act irreducibly on and it also does not induce power automorphisms on
. In particular it does not act trivially. Since PU is a proper subgroup of our minimal counterexample G, it follows that
. But
is Q-invariant, because Q is abelian. This is a final contradiction with regard to (2). □
Lemma 4.5.
Suppose that q is odd and that G is an L9-free -group. Suppose further that P is a normal Sylow 2-subgroup of G such that
is hamiltonian and let
. Then one of the following holds:
G is nilpotent or
and there exists a group I of order at most 2 such that
and Q is a cyclic 3-group. Moreover
.
Proof.
We suppose that G is not nilpotent.
Then Q is not normal in G and Lemma 4.3 implies that Q is cyclic. Furthermore, P is L9-free and hence it is modular by Lemma 4.2. Since is hamiltonian, Theorem 2.3.1 of [Citation7] provides subgroups
such that
, I is elementary abelian and
.
We recall that the automorphism group of Q8 is isomorphic to . Thus, if
is Q-invariant, but not centralized by Q, then
and
. It follows that Q is a cyclic 3-group and
.
We conclude that our assertion holds if I = 1. Now suppose that . We recall that
and therefore
. Since
, it follows that
, and then
is not nilpotent because G is not. Furthermore,
is L9-free,
and
is elementary abelian of order 4. In particular
is elementary abelian, hence it is a non-hamiltonian 2-group of order at least 8. Lemma 4.4 states that
acts irreducibly on
or induces power automorphisms on it. The second case is not possible by Lemma 1.3.
Hence acts irreducibly on
and, by Lemma 4.4, we see that
is a cyclic 2-group. Since
and
, we have that
is
-invariant and
is a non-cyclic complement of
in
. This implies that
is cyclic and elementary abelian at the same time. Thus
has order 2 and with Lemma 1.1 we deduce that
is elementary abelian of order 4. Then we deduce that
and then
. Moreover, since
, it follows that
. In conclusion,
is a subgroup of order at most 8 admitting an automorphism of odd order that centralizes
. It follows that
and then, together with the fact that
, our assertions follow. □
Definition 4.6.
Suppose that Q is a cyclic q-group that acts coprimely on the p-group P. We say that the action of Q on P avoids L9 (and we indicate more technical details by writing “of type (
)”) if and only if one of the following is true:
(std)Every subgroup of Q acts irreducibly or by inducing (possibly trivial) power automorphisms on the elementary abelian group .
(cent)Every subgroup of Q acts irreducibly or trivially on the elementary abelian group , P is abelian and
is a nontrivial cyclic 2-group.
(hamil) , where I is a group of order at most 2, and Q is a cyclic 3-group such that
.
Lemma 4.7.
Suppose that p is an odd prime and that G is an L9-free -group. Let further P be a normal Sylow p-subgroup of G and let
be isomorphic to Q8 and such that
is elementary abelian.
Then and Q acts faithfully on P.
Proof.
We set . If
, then
and we consider
. Then
is an L9-free
-group,
is a normal Sylow p-subgroup of
and
. Since
is neither cyclic nor isomorphic to Q8, Lemma 4.3 is applicable and we see that
is nilpotent. But then G is also nilpotent, contrary to our hypothesis that
. Thus
is not normal in G and Q acts faithfully on P. Now, for all
of order 4, we apply Lemma 4.4 on
to deduce that
either induces power automorphisms on
or acts irreducibly on it. Theorem 1.5.1 of [Citation7] states that
is abelian, but
is not, which means that we may choose y such that y does not induce power automorphisms on P. In particular P is not cyclic of prime order. Moreover p is odd and therefore 4 divides
, and Satz II 3.10 of [Citation3] yields that
. It follows that
. More precisely, as
, the result implies that
, and then the proof is complete. □
Definition 4.8.
Suppose that acts coprimely on the p-group P. We say that the action of Q on P avoids L9 if and only if
and Q acts faithfully on P.
Lemma 4.9.
Suppose and that P is a p-group on which Q acts avoiding L9. Then P is elementary abelian,
inverts P, and every subgroup of Q of order at least 4 acts irreducibly on P.
Proof.
Since a cyclic group of order p2 has an abelian automorphism group by 2.2.3 of [Citation5], it follows that P is elementary abelian. If is a cyclic subgroup of P, then
and therefore R does not admit an automorphism of order 4. Additionally,
is Q-invariant and, since Q acts faithfully on P, we see that
. Furthermore, Lemma 1.1 gives that
because P is abelian. Moreover, Q has rank 1, and then it follows that Q acts faithfully on
. This implies that
and consequently
. Hence 8.1.8 of [Citation5] states that
inverts P. In particular
inverts every cyclic subgroup R of P.
In addition, these arguments show that every subgroup U of order 4 of Q does not normalize any nontrivial proper subgroup of the elementary abelian group P. This means that U acts irreducibly on P. □
Corollary 4.10.
Suppose that and that G is an L9-free {p, q}-group such that
is normal in G and
.
Then either G is nilpotent and P and Q are modular or Q is a batten and it acts on P avoiding L9.
In particular, if G is not nilpotent, then Q is isomorphic to Q8 or cyclic and is elementary abelian or isomorphic to Q8, where in the second case q = 3.
Proof.
By hypothesis G is L9-free, hence P and Q are, too. Then Lemma 4.2 implies that P and Q are modular.
Suppose that G is not nilpotent. Then Lemma 4.3 applies: Q is cyclic or isomorphic to Q8 and hence it is a batten. Moreover Lemma 2.2 of [Citation9] states that is a hamiltonian 2-group or elementary abelian. In the first case Lemma 4.5 gives the assertion. In the second case our statement follows from Lemmas 4.4 and 4.7. □
Lemma 4.11.
Let Q be a nilpotent batten that acts on the p-group P avoiding L9, and suppose that U is a subgroup of Q.
Then U induces power automorphisms on P or it acts irreducibly on .
Proof.
First suppose that . Then Lemma 4.9 implies that every subgroup of order at least 4 of Q, and in particular Q itself, acts irreducibly on
. Moreover, the involution of Q inverts P by Lemma 4.9, and then the statement holds.
Next we suppose that Q is cyclic. Then Definition 4.6 gives the assertion unless the action of Q on P avoids L9 of type (hamil). In this case every proper subgroup of Q centralizes P, while Q acts irreducibly on . □
Next we investigate groups of order divisible by more than two primes. This needs some preparation.
Lemma 4.12.
Suppose that P and R are distinct Sylow subgroups of G, that is cyclic and that it normalizes P and R, but does not centralize them. Suppose further that R normalizes every Q-invariant subgroup of P. If
, then G is not L9-free.
Proof.
We suppose that . Then E is a normal subgroup of G because Q is abelian.
We claim that G/E is not L9-free, and for this we may suppose that E = 1. Then acts faithfully on P and R. Now we need a technical step before we move on:
There are a Q-invariant subgroup D of P and elements such that
and
. (*)
Since Q is cyclic, there is some such that
. Let
be such that
does not centralize x0.
Then by Lemma 1.1, and for all integers n we have the following:
iff
iff
. It follows that
and
.
Now we set and
. Then D is Q-invariant and we have that
and
. If we set
, then
and we can use the information from the end of the previous paragraph:
and
. This concludes the proof of (*).
We use (*) and its notation and, similarly, we find a Q-invariant subgroup R0 of R and an element such that
and
.
We set ,
and
, and we claim that
is isomorphic to L9. The properties L9 (i) and L9 (iii) of Lemma 3.5 follow from the choice of D, since h and Q normalize D.
For L9 (ii) we first note that and
, since
and hence
. Next, Lemma 1.4 (b) yields that
. Part (d) of the same lemma shows that
, as
.
For all we calculate that
by Lemma 1.4 (d). Thus L9 (iv) of Lemma 3.5 is true.
We moreover have that and
. Finally
by Lemma 1.4 (b).
Using Lemma 3.5 we conclude that G/E is not L9-free, and hence G is not L9-free. □
Corollary 4.13.
Suppose that and r are pairwise distinct primes and that G is a directly indecomposable L9-free {p, q, r}-group. Suppose further that
and
are normal in G and let
.
Then Q is cyclic and .
Proof.
Since G is directly indecomposable, we see that Q acts non-trivially on both P and R. Moreover, PQ and RQ are L9-free by hypothesis, and then we conclude that Q is cyclic or isomorphic to Q8.
In the first case, our assertion follows from Lemma 4.12, and in the second case, we choose a maximal subgroup Q1 of Q. Then Q1 acts irreducibly on P and R by Lemma 4.9, and the same Lemma shows that inverts P and R. Thus Q1 acts on P and on R avoiding L9, respectively, and it acts faithfully. This contradicts Lemma 4.12. □
We explain another example where a subgroup lattice contains L9.
Lemma 4.14.
Suppose that p, q and r are pairwise distinct primes and that G is a {p, q, r}-group. Suppose further that is normal in G and that
and
are cyclic groups such that
. Suppose that
and
.
If R acts irreducibly on P, but non-trivially, and if is elementary abelian, then G is not L9-free.
Proof.
We first remark that G is soluble, because . We will construct the lattice L9 in L(G) using Lemma 3.5. For this we set
and
. Then
and we see that L9 (i) is true.
Next, we recall that by hypothesis. Assume that
. Then Q, which normalizes
, must centralize it, and then Lemma 1.1 gives a contradiction.
Therefore . As a consequence, we find
such that
, and then we set
and
. Now
and
. We set
. In addition, since R acts irreducibly, but non-trivially on P, it follows that
. Then the fact that
gives that
.
Lemma 1.4 (b) shows that , and now we recall that
. Moreover, R acts irreducibly on P, and then Part (e) of the same lemma yields the following:
. We conclude that L9 (ii) holds.
For L9 (iii) we set and
. Then we note that
.
Assume for a contradiction that has odd order for some
.
In both cases X is a -Hall subgroup of the soluble group G = PRQ and therefore
. It follows that Q normalizes Rc and then that
and Q normalize R.
Since is R-invariant and R acts irreducibly, but non-trivially on P, we conclude that
. Thus Sylow’s theorem provides some
such that
. Now
. In addition
and
by hypothesis. Together this gives that
, by Dedekind’s modular law. Altogether we have that
. We recall that
, and then
. But we chose
and now
, whence
. In other words,
, and this means that
and
. We recall that
and that
is elementary abelian by hypothesis. Then Lemma 1.1 implies that
, and this contradicts the fact that
.
It follows that X has even order and since Rc acts irreducibly on P, we conclude that . This implies that
and then L9 (iv) holds for
.
We finally set for some
. Then
and Lemma 1.4 (b) and (c), together with our hypothesis, show that
. Thus we have that
. We further see that
and
. Since
and
by Lemma 1.4 (d), we finally obtain L9 (v).
Altogether Lemma 3.5 gives the assertion. □
Proposition 4.15.
Suppose that , and r are pairwise distinct primes and that G is a non-nilpotent L9-free {p, q, r}-group with normal Sylow p-subgroup P. Suppose further that
and
are not normal in G, that
and
.
Then RQ is a batten, P is elementary abelian of order pr, R and Q act irreducibly on P and induces non-trivial power automorphisms on P.
Proof.
We proceed in a series of steps.
(1) The groups PQ and PR are not nilpotent, Q is cyclic and or
. In addition
and
.
Proof. By hypothesis R is not normal in G, but Q normalizes R. Hence and in particular PR is not nilpotent. But PR is L9-free, because G is. Moreover, RQ is non-nilpotent and L9-free, again by hypothesis. Then Corollary 4.10 implies that R and Q are battens, that R acts on P avoiding L9 and that Q acts on R avoiding L9. More specifically, R and Q are cyclic or isomorphic to Q8, and
as well as
are elementary abelian or isomorphic to Q8.
It follows that or that R is cyclic of order r. In both cases Lemma 1.1 yields
and the avoiding L9 action of Q on R gives that Q is cyclic.
In addition and then the Three Subgroups Lemma (see for example 1.5.6 of [Citation5]) implies that
. Thus
.
If it was true that , then
would centralize P. But this is a contradiction. □
(2) and
.
Proof. Since by (1), we can apply Corollary 4.10 to PR, and this shows that R acts on P avoiding L9. Otherwise (1) implies that
and then Definition 4.8 gives that R acts faithfully on P. If
, then R acts faithfully of P because
. In both cases we see that
, and then the last statement of (1) implies that
. Then
centralizes P and R, and Q is cyclic by (1), and therefore it follows that
. □
(3) Z(G) = 1 or .
Proof. We suppose that p = 2. Let be the natural homomorphism. We show that
satisfies the hypotheses of our lemma.
From (1) we see that none of the groups P, Q or R is contained in Z(G). We even have that by (1). In particular
and
is L9-free. We see that
is a normal Sylow p-subgroup of G and that
and
are such that
. Let
. If
, then
and X is a characteristic Sylow subgroup of XZ(G) and hence normal in G. This is a contradiction.
We deduce that all hypotheses of the lemma hold for and that
. Therefore, if
, then the minimal choice of G implies that
induces non-trivial power automorphisms on the elementary abelian group
. Then Lemma 1.3 yields that
. □
(4) , and the groups
and
are elementary abelian.
Proof. As PR and PQ are not nilpotent by (1), Corollary 4.10 implies that X acts on P avoiding L9 and that is elementary abelian or isomorphic to Q8 for both
.
Assume for a contradiction that is isomorphic to Q8 for some
. Then X acts on P of type (hamil). Hence we obtain a group I of order 1 or 2 such that
. It follows that
is a {2, 3}-group. But this is impossible because Q and R both act coprimely and non-trivially on P by (1), and p = 2, q and r are pairwise distinct.
We conclude that and
are elementary abelian, and then it follows that P is abelian, by Lemma 4.4, applied to PR.
Assume for a further contradiction that . As R avoids L9 in its action on P, it follows from Lemma 4.9 that R is not isomorphic to Q8. Now (1) yields that R is cyclic and we may apply Lemma 4.4 to PR, because
is elementary abelian. It follows that
is a cyclic 2-group and thus q and r are odd.
Furthermore is normalized by Q, because Q normalizes P and R. Since q is odd, it follows that
is centralized by Q. We recall that P is abelian, and then (3) yields that
. This is a contradiction.
Altogether and Lemma 1.1 gives that
is elementary abelian. □
(5) and Q acts on P avoiding L9 of type (std).
Proof. Since PQ is not nilpotent and Q is cyclic by (1), Corollary 4.10 implies that the action of Q on P avoids L9. But P is elementary abelian by (4), and therefore the action is not of type (hamil).
We assume for a contradiction that . Then it follows that PQ is not of type (std). We consequently have type (cent) and we see that
is a cyclic 2-group. In particular p = 2, and thus q and r are odd. Then
by (1).
Next we claim that G satisfies the hypotheses of Lemma 4.14.
First, q and r are pairwise distinct odd primes and G is a finite -group. From above, (1) and our assumption we see that
is normal in G and that
and
are cyclic groups such that
. We have shown that R has order r.
We recall that is a cyclic 2-group (first paragraph). As r is odd and
by (4), we see that
is not R-invariant. But
is R-invariant, and now the irreducible action of R on P and the fact that
show that
. Then (2) and (4) imply that
.
By (4) P is an elementary abelian 2-group, and by (1) and (4). In particular R does not induce power automorphism on P by Lemma 1.3. As
by (4), we deduce that R acts irreducibly on P (using Lemma 4.4). Finally, (1) and (4) yield that
is elementary abelian.
All hypotheses of Lemma 4.14 are satisfied now, and we infer that G is not L9-free. This is a contradiction.
Thus and we deduce that the action of Q on P is not of type (cen). It remains that Q acts on P avoiding L9 of type (std). □
(6) R and Q act irreducibly on P. If induces power automorphisms on P, then X centralizes R.
Proof. We recall from (2) that and
. Consequently
and
is isomorphic to a subgroup of
. Since RQ is not nilpotent, we see that
is not nilpotent.
The group P is an elementary abelian p-group by (4) and hence, if induces power automorphisms on it, then it follows that
, see page 177 of [Citation3]. We denote this fact by (*). Then we deduce that
(see above) and therefore
by Lemma 1.1.
In addition, the fact (*) shows that neither R nor Q induces power automorphisms on P. It follows from (5) and Definition 4.6 (std) that Q acts irreducibly on P. Furthermore (1), together with Corollary 4.10, yields that R acts on P avoiding L9. Since by (4), this action has type (std) or (hamil). In the first case, the irreducible action follows from Definition 4.6 (std), and in the second case, it follows from Lemma 4.9. □
(7) induces nontrivial power automorphisms on P.
Proof. We set and we assume that Q0 acts irreducibly on P. Then II 3.11 of [Citation3] implies that
is isomorphic to a subgroup of the multiplicative group of some field of order
, and it follows that RQ is cyclic. This is a contradiction.
Thus (6) and Lemma 4.11 show that Q0 induces power automorphisms on P. Since R normalizes every Q-invariant subgroup of P by (7), we see from Lemma 4.12 and (2) that . □
(8) and
.
Proof. We recall that Q is cyclic, by (1). Moreover , whence
and therefore we may choose
such that
.
By (5) and Lemma 4.11, it follows that every subgroup of either acts irreducibly on P or induces power automorphism on it (in particular normalizing every subgroup of P). Then
satisfies (b) of Lemma 3.1 in [Citation8], which implies that it satisfies one of the possibilities 3.1 (i)–3.1 (iii). By (5) and the choice of y, we see that 3.1 (i) is not true. Further (7) provides some
that induces a power automorphism of order q on P. This implies that q divides p – 1 and therefore
satisfies (ii) of Lemma 3.1 (b) in [Citation8]. It follows that
and that, if k is the largest positive integer such that qk divides p – 1, then y induces an automorphism of order
on P. We conclude that
, because Q is cyclic. Finally, we deduce that o(y) is uniquely determined, that
and that
. □
By (6), we see that R and Q act irreducibly on P, and (1) gives that Q is cyclic. Then (8) and (7) say that induces nontrivial power automorphisms on P. In addition P is elementary abelian by (4), and it has order pr by (8). If
, then R is a cyclic group of order r and RQ is a batten. In particular G satisfies the assertion of our lemma.
But G is a counterexample, and then it follows that and r = 2. Then (4) yields that PR fulfills the hypothesis of Lemma 4.7, and consequently
by (8). This is our final contradiction, because
. □
Definition 4.16.
Suppose that B is a non-nilpotent batten that acts coprimely on the p-group P. We say that the action of B on P avoids L9 if and only if and if one of the following occurs:
(Cy) and Q acts on P avoiding L9 for every Sylow subgroup Q of B different from
or
(NN)P is elementary abelian of order and the Sylow subgroups of B act irreducibly on P, while Z(B) induces power automorphisms on P.
As in Definition 4.6, we specify the type of the L9-avoiding action by writing that “B acts on P avoiding L9 of type (
)”.
Lemma 4.17.
Let B be a batten that acts non-trivially and avoiding L9 on the p-group P. Then the following hold:
If
, then p = 2.
Either
, where
is elementary abelian and
is cyclic, or
, where I is a group of order at most 2 and
.
is centralized by every automorphism of P of order coprime to p that leaves
invariant.
Proof.
If , then Definition 4.8 and Lemma 4.9 imply that P is elementary abelian and that B acts irreducibly on it. We conclude that
is elementary abelian and we deduce from Lemma 1.1 that
. Hence, in this case, all statements of our lemma hold.
Now suppose that B is not nilpotent and that it acts of type (NN). Then Definition 4.16 states that, once more, P is elementary abelian and B acts irreducibly on it. Again we see that is elementary abelian, and as before all statements hold.
Next we suppose B is not nilpotent and that it acts of type (Cy), or that B is cyclic. In the first case B has a cyclic Sylow subgroup Q that acts on P avoiding L9 such that and
by Definition 4.16. In the second case we set
.
Then, in both cases, Q is a cyclic group that acts on P avoiding L9 such that and
. If Q acts of type (std), then
is elementary abelian by Definition 4.6. Again we deduce the statements of our lemma.
Suppose that Q acts of type (cent). Then Definition 4.6 yields that is elementary abelian, that P is abelian and that
is a cyclic 2-group. In particular
by Lemma 1.1. It also follows that
is centralized by every automorphisms of P of odd order that leaves
invariant. These are the statements of our lemma.
Finally, suppose that Q acts of type (hamil). Then Definition 4.6 yields that and
, where I is a group of order at most 2. In particular statement (a) is true. Moreover, we deduce that
, where
is cyclic of order 2. In particular
is centralized by B and by every automorphisms of P. We conclude that
is elementary abelian of order at most 4 and that every automorphism of P centralizes a cyclic subgroup of order 2. This implies (b). □
5 Avoiding L9
We now work toward a classification of arbitrary L9-free groups, and therefore we need to understand in more detail the group structures that appear when “L9 is avoided” in the sense of the previous section.
Definition 5.1.
Suppose that K is a batten group that acts coprimely on the p-group P. We say that the action of K on P avoids L9 if and only if and every batten of K either centralizes P or avoids L9 in its action on P.
Lemma 5.2.
Let K be a batten group that acts coprimely on the p-group P avoiding L9. Suppose further that and
such that
. Then
acts on P avoiding L9. In particular
is a batten group.
Proof.
By induction we may suppose that K is a batten and that either and L is a maximal subgroup of K or that L0 is a minimal normal subgroup of K = L. Thus either
has order q or
. Since
, we first remark that
induces automorphisms on P.
If , then K acts faithfully on P by Definition 4.8. Thus
and it follows that L is a cyclic group of order 4. Thus Lemma 4.9 yields that
inverts P and that L acts irreducibly on the elementary abelian group
. Then we see that
acts on P avoiding L9 of type (std).
Next suppose that K is cyclic. Then is cyclic. If K acts of type (std) on P, then L and every subgroup of L act irreducibly or via inducing power automorphisms on the elementary abelian group
. Since
and power automorphisms are universal, by Lemma 1.5.4 of [Citation7], it follows that
. Moreover, the action of L on P is equivalent to that of
, and then it follows that
acts on P avoiding L9 of type (std).
If K acts on P of type (cent), then L and all its subgroups act irreducibly or trivially on the elementary abelian group . Again the fact that
implies that
, and then
by Lemma 1.1. Since the action of L on P is equivalent to that of
, it follows that
acts on P avoiding L9 of type (cent).
We suppose now that K acts of type (hamil). Then K is a cyclic 3-group and has order 3. It follows that L = K. But again, the action of
on P is equivalent to the action of K on P, which means that it has type (hamil).
We finally suppose that K is a non-nilpotent batten. Let R be a Sylow subgroup of K such that . Suppose first that
is a q-group. Then our choice of L and L0 implies that
. If K acts on P of type (Cy) in this case, then it follows that
is cyclic and that it acts on P avoiding L9, according to Definition 4.16. Otherwise, if K acts of type (NN), then
and then
. It follows that L = R acts irreducibly on the elementary abelian group P, whence
. In addition
and all of its subgroups induce power automorphism on P. Altogether the cyclic group
acts on P of type (std).
Now we suppose that does not have prime power order. Then
. Now if
is nilpotent, then
and therefore
. It follows from Lemma 2.5 that
. We have already proven that R acts on P avoiding L9, and then Z(K) also acts on P avoiding L9. In addition
either centralizes P or it acts irreducibly on the elementary abelian group
. Since
has prime order, it follows that the cyclic group
acts on P avoiding L9 of type (std). Altogether
acts on P avoiding L9.
Finally, suppose that is not nilpotent. Then L is not nilpotent and hence Lemma 2.7 implies that L = K. We conclude that
. Since
by Definition 4.16, it follows that L0 is a proper subgroup of Z(K) and that
is a non-nilpotent batten. If
, then our investigation above imply that the cyclic group
acts on P avoiding L9, and then
acts on P avoiding L9. Otherwise
is elementary abelian of order
, moreover
and
act irreducibly on P. At the same time
induces power automorphisms on P. Altogether
acts on P avoiding L9 of type (NN). □
Lemma 5.3.
Let K be a batten group that acts coprimely on the p-group P avoiding L9. Then the following assertions are true:
If
, then
or
.
is elementary abelian or isomorphic to Q8.
Proof.
Let be such that
. Then L is a batten group by Lemma 2.7 and therefore there is a batten B of L such that
. Assume for a contradiction that
. Then the fact that
implies that
by Lemma 1.1. In addition B avoids L9 in its action on P, by Lemma 5.2. Since B is a batten of L, it is characteristic in L, and therefore
. In particular
is K-invariant and hence it is centralized by K by Lemma 4.17 (c). This implies that
. Finally
, which is a contradiction. In particular (a) is true.
Together with Lemma 4.17 (b), the statement in (b) follows from (a). □
Lemma 5.4.
Let K be a batten group that acts on the p-group P avoiding L9, and suppose that H is a subgroup of K. Then H centralizes P or .
Moreover, H induces power automorphisms on P or it acts irreducibly on .
Proof.
Let . If H centralizes P, then it induces power automorphisms on P. We may suppose that
. Then H has a q-subgroup Q such that
. Therefore, if
, then we have that
by Lemma 5.3 (a). Then the fact that
yields that
.
Assume for a contradiction that . Then Lemma 2.8 implies that K has a non-nilpotent batten B such that
. We moreover deduce that
and
by Lemma 5.3 (a), because
. Since the action of K on P avoids L9, the action of B also does. If B acts of type (Cy), then we obtain the contradiction that
. Thus B acts of type (NN) and in particular Q acts irreducibly on P. But this is impossible as well. It follows that
.
Assume for a further contradiction that neither induces power automorphisms on P nor does it act irreducibly on
. Then there is a batten B of H that neither induces power automorphisms on P nor does it act irreducibly on
. Similarly to the arguments above, we deduce that
, and Lemma 5.2 gives that B avoids L9 in its action on P. Therefore Lemma 4.11 yields that B is not nilpotent. From Definition 4.16 we further see that B does not act of type (NN), and thus B acts of type (Cy) on P. Consequently
and B has a cyclic Sylow subgroup Q such that
and Q acts on P avoiding L9. Again we have
and Lemma 4.11 gives that Q induces power automorphisms on P or acts irreducibly on
. In the first case
induces power automorphism on P and in the second case B acts irreducibly on
. This is a contradiction. □
Lemma 5.5.
Let K be a batten group that acts on the p-group P avoiding L9, and suppose that H is a subgroup of K that acts non-trivially on .
Then and
.
Proof.
From Lemma 5.4 we see that . In addition
for every q-subgroup Q of H and every prime q. Let Q be a q-subgroup of H for some prime q such that
. Then Q is a batten by Lemma 2.7, and it acts on P avoiding L9 by Lemma 5.2. If
, then K centralizes
by Lemma 4.17(c). Thus
, and this gives that
. If Q is not a normal subgroup of K, then Lemma 2.8 provides a non-nilpotent batten B of K such that
. Since the action of K on P avoids L9, the action of B also does. If B acts of type (NN), then Q acts irreducibly on P and therefore
. Again we deduce that
. If B acts of type (Cy), then
and hence
. But now B is a normal subgroup of K, and then Lemma 4.17(c) gives that
. As above we deduce that
.
Finally, suppose that is H-invariant, but not centralized by H, and set
. Assume for a contradiction that H0 does not centralize P. Then we deduce, as above, that
. This is a contradiction, because H does not centralize R. □
Corollary 5.6.
Let K be a batten group that acts non-trivially and avoiding L9 on the p-group P. Then the following hold:
If
, then p = 2.
If
, then
is elementary abelian.
If K induces power automorphisms on P, then
is elementary abelian of odd order. In particular
in this case.
Proof.
Let B be a batten of K that does not centralize P. Then Lemma 5.5 implies that . Thus Part (a) and (b) of Lemma 4.17 yield the statements (a) and (b) of our lemma. For Part (c) we suppose that K induces power automorphisms on P. Then P is not an elementary abelian 2-group by Lemma 1.3. If
, then our assertion holds by (b). Otherwise p = 2 by (a), and then Lemma 1.3 implies that
is neither elementary abelian nor isomorphic to Q8, contradicting Part (b) of Lemma 5.3.
For the final comment we just use that p is odd and then apply (a). □
Lemma 5.7.
Let B be a batten that acts on the p-group P avoiding L9. Let be B-invariant and
.
Then B avoids L9 in its action on .
Proof.
Since B centralizes R0, the action of B on is well-defined.
We first suppose that . Then Lemma 4.9 yields that B acts irreducibly on P, and then it follows that R = P and
. Thus our assertion is true in this case.
Next suppose that B is cyclic. If B acts of type (std) on P, then . If B acts irreducibly on P, then again P = R and there is nothing left to prove. Otherwise B and all of its subgroups induce power automorphisms on P, and hence on
as well. It follows that B also acts of type (std) on
.
Suppose now that B acts of type (cent). Then, since B does not centralize R and B acts irreducibly on , it follows that
. Moreover P is abelian and then we use the fact that
. This gives that
, where
is a cyclic 2-group. Since every subgroup of B that does not centralize P acts irreducibly on
in this case, there are two possibilities for the action of B on
: If
, then B acts of type (cent), and otherwise it acts of type (std).
Suppose now that B acts of type (hamil). Then the cyclic 3-group B acts irreducibly on , by Lemma 5.4, and we see again that
. It follows that
, where I is a group of order at most 2, and then
. We remark that
.
If , then
for some group J of order
. Thus B acts on
of type (hamil) in this case.
Otherwise we have that and therefore
is elementary abelian of order 4 or 8. Moreover B acts irreducibly on
, which is a group of order 4. In addition every proper subgroup of B centralizes
. Consequently, if
, then B acts on
of type (std) or of type (cent).
The final case is that B is not nilpotent, and we suppose that B acts of type (NN) on P. Then Definition 4.16 yields that B acts irreducibly on P. Hence there is nothing left to prove.
Suppose that B acts of type (Cy). Then we choose a Sylow subgroup Q of B such that . Then
and Q acts on P avoiding L9 in such a way that
does not centralize P. Then Lemma 5.5 yields that
does not centralize R. In particular, we have that
. In addition Q acts on
avoiding L9, by our arguments above. Altogether B acts on
avoiding L9 of type (Cy) in this final case. □
6 The first implication
We now investigate the general case.
Proposition 6.1.
Let G be a finite L9-free group. Then G = NK, where N is a nilpotent normal Hall-subgroup of G with modular Sylow subgroups and K is a batten group. Moreover, for all , every batten of K acts on
avoiding L9 or it centralizes
.
Proof.
We first remark that G is L10-free, whence Theorem A
of [Citation9] implies that G is soluble. Furthermore, Corollary C of [Citation9] provides normal Hall-subgroups N and M of G such that and such that N is nilpotent, M/N is a 2-group and G/M is metacyclic. We choose N as large as possible with these constraints. From Lemma 4.1 we see that
. We also have that every Sylow subgroup of N is L9-free, and hence it is modular by Lemma 4.2. In addition the Schur-Zassenhaus Theorem (see for example 3.3.1. of [Citation5]) provides a complement K of N in G.
(1) If RQ is a non-nilpotent Hall {r, q}-subgroup of K, where R is a normal Sylow r-subgroup of RQ and , then RQ is a batten. For all
, the group RQ centralizes
or acts on it avoiding L9.
Proof. If there is some such that
, then we set
. Otherwise the maximal choice of N implies that R is not a normal subgroup of K. Then, using the solubility of G, we find a prime
and a normal s-subgroup T of K such that
(see 5.2.2 of [Citation5]). Then
because
and
. In this case we set
.
In both cases and q are pairwise different primes and PRQ is a non-nilpotent {p, r, q}-subgroup that satisfies the hypothesis of Proposition 4.15. For this we note that
and
. Since
, the assertion in (1) follows. □
(2) Every Sylow subgroup S of K is a batten, and for all it is true that S centralizes
or acts on
avoiding L9.
Proof. Let S be a Sylow subgroup of K. If S centralizes N, then the choice of N provides some Sylow subgroup R of K such that RS is not nilpotent. Then (1) implies that RS is a batten, then that S is cyclic and hence that S is a batten.
Let . Then
is an L9-free {p, q}-group for some prime q. Hence Corollary 4.10 implies the assertion. □
(3) K is a batten group.
Proof. Let be such that there is some
such that
, where
and B is not a direct product of nontrivial subgroups of coprime order. In particular B is a Hall subgroup of K. If B is nilpotent, then B is a Sylow q-subgroup of K for some prime q. In this case (2) implies that B is a batten.
Assume for a contradiction that B is not a batten of K. Then B is not nilpotent and therefore (1) yields that is divisible by at least three different primes. Since
is L9-free, Lemma 4.1 provides a normal Sylow r-subgroup R of B for some prime
. We remark that R is a normal subgroup of
. In addition B is is not a direct product of non-trivial subgroups with coprime order and hence there are a prime
and some
such that RQ is not nilpotent. Now B is a Hall subgroup of K and thus RQ is a Hall subgroup of K. In particular (1) implies that RQ is a batten and it follows that
and
. We further see, from Definition 4.16 and (1), that for every
with the property
we have that
. Since R is a normal Sylow subgroup of K, the maximal choice of N provides some
such that R does not centralize
. In particular we have that
.
Let and let S be a Sylow s-subgroup of B such that QS = SQ. Such a subgroup exists by Satz VI. 2.3 in [Citation3]. If S does not centralize R, then RS is not nilpotent and therefore our arguments above show that
. This is impossible because
. Consequently
.
Since B is directly indecomposable, we conclude that SQ is not nilpotent. But SQ is a Hall subgroup of B and then it is a Hall subgroup of K. In particular (1) yields that SQ is a batten. From the fact that we conclude that
, and then
and
. In addition
. But
is an L9-free group, and this situation contradicts Corollary 4.13. □
We conclude that, by construction, G = NK, where N is a nilpotent normal subgroup of G with modular Sylow subgroups and K is a batten group, by (3), such that for all it is true that every batten of K centralizes
or acts on
avoiding L9, by (1) and (2). □
The converse of Proposition 6.1 is false, as can be seen in the following example and subsequent lemma.
Example 6.2.
Let and let
be such that
Then . Hence
is a group.
Moreover, is a nilpotent normal subgroup of G with modular Sylow subgroups. Since
and
, it follows that x and y have coprime order.
Thus is cyclic, which means that it is a batten group.
We see that x and y induce non-trivial power automorphisms on H. Thus every batten of K acts on avoiding L19 of type (std). In addition x2 and y3 induce power automorphisms on J. Since x and y act irreducibly on J, it follows that every batten of K acts on
avoiding L19 of type (std), too.
Altogether G satisfies the conclusion of Proposition 6.1.
On the other hand we observe that and that
.
If centralizes
or
, then g = 1. Thus the following lemma yields that G is not L9-free.
Lemma 6.3.
Let H be a non-trivial abelian group where all non-trivial Sylow subgroups are non-cyclic elementary abelian, let L be a cyclic group inducing power automorphism on H such that and let
be an abelian group admitting L as a group of automorphisms such that the action of L on
avoids L9 for every
and such that
.
Let . Suppose that
and, for all
, suppose that g centralizes neither
nor
.
Then is not L9-free.
Proof.
We first set and
. Since none of the groups L1 nor L2 is centralized by any element of
, it follows that L1 and L2 are both non-trivial.
For every odd prime there is an elementary abelian subgroup Hp of H that has order p2. In particular there are elements ap and bp of H such that
.
We set and
.
These elements are well-defined (in the sense that the ordering of the primes does not matter) because H is abelian. For every , we further see that L acts on
avoiding L9. Since J is abelian, we deduce from Lemma 5.3(b) that
is elementary abelian. In addition L acts irreducibly on
or it induces power automorphisms on
, by Lemma 5.4. We choose
. Then L acts irreducibly on
. Next we choose
such that
. Then
and
. Thus our hypothesis implies that
and
. Altogether we have that
. Moreover, Lemma 5.5 yields that
.
For every we choose
, and then we set
.
Since J is abelian, it follows that . Next we set
. Then our previous arguments show that
(*).
We will construct a subgroup lattice L9 using Lemma 3.5. For this we set and
.
For every we have
and so
. This implies that
(
) and that
. We conclude that
.
If , then
and therefore
(
). Then it follows that
. In particular
and hence (L9(i)) of Lemma 3.5 holds.
Next we set and
. Then we have that A contains the subgroups S, T and D (
). In addition, if
, then we see that
, because o(a) is odd by construction. Since
, it follows that
, by Lemma 1.1. The group L1 induces power automorphisms on H, which means that it normalizes
. Together with Part (d) of Lemma 1.4 we conclude that
In particular, since D centralizes x, it follows that . Moreover, we have that
and therefore we conclude that
as well. Next Lemma 5.5 gives that
by (
). Furthermore
is a π-group for all
, whence
by Part (b) of Lemma 1.4. Altogether Dedekind’s modular law gives that
for all
. We conclude that
, that
and that
. With all these properties, we see that (L9(ii)) of Lemma 3.5 is true.
Now we set and
. Then
and
by Dedekind’s modular law and by Part (b) of Lemma 1.4. Hence (
) implies that (L9(iii)) of Lemma 3.5 holds.
Let and let
. Then X contains a π-Hall subgroup as well as a
-Hall subgroup of HJL. Since HJL is soluble, there is a
-Hall subgroup K of X such that
and some
such that
. It follows that
by Lemma 1.4(b). The hypothesis of our lemma yields that g = 1 and therefore
. From there we obtain some
such that
and hence
by Lemma 1.4(b). This forces
, and then
, because H and J have coprime order and centralize each other. Altogether
and
are subgroups of X, and we conclude that
. Thus
We set in order to obtain Part (L9(iv)) of Lemma 3.5. Moreover, Dedekind’s law and Part (a) of Lemma 1.4 gives that
We set . Then
by Lemma 1.4(b), because
.
In a similar way we obtain that and therefore
.
We further calculate that
and similarly
. Therefore
.
Finally (*) and Part (d) of Lemma 1.4 yield that
Altogether satisfies every condition of Lemma 3.5, which means that it is isomorphic to L9. □
The previous lemma and Lemma 4.12 motivate the following definition:
Definition 6.4.
Here we define a class of finite groups, and each group in
has a type.
We say that has type (N, K) if and only if the following hold:
(1)
, where N is a normal nilpotent Hall subgroup of G with modular Sylow subgroups and K is a batten group.
(2)If
, then every batten of K centralizes
or it acts on it avoiding L9.
(3)For all Sylow subgroups Q of K and all distinct Sylow subgroups P and R of N that are not centralized by Q, we have that
.
(4)Suppose that
is abelian, that its nontrivial Sylow subgroups are not cyclic and that
is cyclic and such that
. Let
be L-invariant and abelian, suppose that
and
, and set
.
Then there is some that centralizes
or
.
Theorem 6.5.
Let G be a finite L9-free group. Then .
Proof.
From Lemma 6.1 we see that G = NK and that (1) and (
2) are satisfied.
For (3) we let Q be a Sylow subgroup of K and we let P and R be distinct Sylow subgroups of N that are not centralized by Q. Since N is a nilpotent normal Hall subgroup of G, it follows that
and that Q normalizes P and R. Then
is directly indecomposable and L9-free, whence Corollary 4.13 gives that
.
Finally, we look at (4) and we assume that it is not true. Then there is an abelian subgroup H of N such that the nontrivial Sylow subgroups are not cyclic, and we find a cyclic group
such that
and a nontrivial L-invariant abelian subgroup J of N such that
and
. Let
. Then we have, for all
, that g centralizes neither
nor
for
.
We note that P does not centralize . Then we find a prime
such that a Sylow q-subgroup Q of L does not centralize P. Using Lemma 5.2, we see that Q acts on
avoiding L9 and then Lemma 5.7 yields that Q acts non-trivially on P, and avoiding L9. Now we may apply Corollary 5.6: Since L induces power automorphisms on P, Part (c) shows that P is elementary abelian. Then the hypotheses of Lemma 6.3 are satisfied. It says that
is not L9-free, which is false. We conclude that (
4) holds. □
7 The class ![](//:0)
![](//:0)
Lemma 7.1.
All groups in the class are soluble.
Proof.
Let be of type (N, K). Then N is nilpotent normal Hall subgroup of G and
is a direct product of p-groups or of groups whose order is divisible by exactly two primes. Thus G/N is soluble as well, and it follows that G is soluble. □
Lemma 7.2.
Let be of type (N, K) and
. Then every subgroup of
is normal in N.
Proof.
Let U be a subgroup of . Then
if and only if
for all
, since N is nilpotent. Let
. We note that this implies that
is not centralized by K. In particular there is a batten of K that acts non-trivially on
and avoiding L9. If
is abelian, then
. If
is not abelian, then we apply Lemma 4.17 (b) to a batten B of K that acts non-trivially on
. The first possibility described there implies that
is abelian, which is not the case here. Thus the second possibility holds, and then
, where I is cyclic of order at most 2. We conclude that
is hamiltonian and it follows that
. □
Lemma 7.3.
Let be of type (N, K) and
. Then there is some
such that
.
Proof.
We set . Then M is a normal Hall subgroup of X, because N is one of G. Then the Schur-Zassenhaus Theorem provides a complement C of M in X, and we notice that C and M have coprime orders. Therefore
. It follows that C is contained in a complement for N in G. Since G is soluble by Lemma 7.1, such a complement is conjugate to K, and thus we find
such that
. The coprime action of K on N yields, together with Lemma 1.1, that
, and therefore
. We notice that
and we let
and
be such that g = yx. Then
and hence
. □
Lemma 7.4.
Let be of type (N, K) and suppose that
. Then U is a group in class
of type
for some
.
Proof.
Lemma 7.3 provides some such that
. By conjugation we may suppose that g = 1 and we set
. Then Lemma 2.7 yields that
is a batten group. Moreover,
is a normal nilpotent Hall subgroup of U with modular Sylow subgroups, by (
1). This means that (
1) holds for U, and now we turn to (
2) and let
. Suppose that B is a batten of K1 that does not centralize
. Then it does not centralize
and therefore Lemma 5.2 implies that
acts on
avoiding L9. Then we may apply Lemma 5.7 to see that B also acts on
avoiding L9.
This gives property of Definition 6.4 for U, and
follows because
and
.
For ( 3) we let Q1 be a Sylow subgroup of K1 and we let
be different primes such that
. We need to prove that
.
First we let Q be a Sylow subgroup of K that contains Q1. Then and therefore
, using Property (
3) for G. In particular, these centralizers cannot both be trivial, and we may suppose that
. Then Q does not act faithfully on
, but the action of Q on
avoids L9. Definition 4.8 immediately gives that
. Then it follows that Q is cyclic, which means that the subgroup lattice L(Q) of Q is a chain, and Q1 is also cyclic.
We assume for a contradiction that . Then Lemma 5.5, with Q1 in the role of H,
in the role of R and
in the role of P, gives that
. Similarly
, and then it follows that
. We recall that
and that Q is cyclic, and now we may suppose that
. This forces
, which is impossible. Thus
and (
3) holds for U as well. □
Lemma 7.5.
Let be of type (N, K) and suppose that S is a normal Sylow q-subgroup of K that centralizes N for some prime q. Let K1 be a Hall
-subgroup of K.
Then G is also of type . In particular, if we choose (N, K) such that
is as large as possible, then
.
Proof.
We show that satisfies (
)–(
) of Definition 6.4, and we first note that K1 is a batten group by Lemma 2.7. The structure of K forces all Sylow subgroups of K to be cyclic or quaternion, more specifically S is cyclic or isomorphic to Q8. This means that S is modular. Since S is a normal Sylow q-subgroup of K and N is a Hall subgroup of G, by hypothesis, it follows that N × S is a Hall subgroup of G where all Sylow subgroups are modular.
By hypothesis and N is nilpotent, hence N × S is nilpotent, too. This is (
).
For () we let B be a batten of K1 and
. We keep in mind that B is not necessarily a batten of K – if it is, then it centralizes
or it acts on it avoiding L9, because of (
) for G.
Now we suppose that B is not a batten of K and that . Then SB is a non-nilpotent batten of K. If
, then SB acts on
avoiding L9, and
. Then Definition 4.16 implies that SB acts of type (Cy) and it follows that B acts on
avoiding L9. Finally suppose that q = p. Then
centralizes
, while B induces power automorphisms on the cyclic group S of order p. Thus SB satisfies (std) of Definition 4.6, and we deduce that B acts on S avoiding L9.
We turn to (). Let Q be a Sylow subgroup of K1 and let P and R be distinct Sylow subgroups of N × S that are not centralized by Q. First we note that Q is a Sylow subgroup of K because K1 is a Hall subgroup of K by hypothesis. Therefore, if
, then we immediately have that
, by (
) in G.
Without loss suppose that , i.e. R = S. Then we recall that Q was chosen not to centralize P and R = S, which means that Q and S cannot come from distinct battens of K, but their product must be a non-nilpotent batten of K. Moreover,
. We obtain from Lemma 2.5 and Definition 4.16 that
does not centralize P and therefore
. This is (
).
Finally, let be such that its nontrivial Sylow subgroups are not cyclic, let
be cyclic and such that
and let
be an L-invariant abelian subgroup of M such that
,
and
.
As K is a batten group, the set contains at most one element. We recall that
, and then it follows that, for every set of primes π, the group S centralizes
or
. In order to prove (
) of Definition 6.4, we may thus suppose that H and J are subgroups of N.
Then shows that
. Hence we apply (
) of Definition 6.4 to G, i.e. to the type (N, K). If
, then we find some
that centralizes
or
.
Altogether, satisfies Definition 6.4.
We now suppose that is as large as possible and we assume for a contradiction that
is a Sylow subgroup of K. Then S is not normal in K, hence there is a non-nilpotent batten B of K such that
. For all
it follows that
, by Lemma 2.5. Then Definition 4.16 yields that B does not act on
avoiding L9, whence B centralizes
. But now
and
is a normal Sylow subgroup of K. This contradicts the maximal choice of N. □
Lemma 7.6.
Let and suppose that
. Then
.
Proof.
By induction we may suppose that M is a minimal normal subgroup of G. We recall that G is soluble by Lemma 7.1, and we let r be prime such that M is an elementary abelian r-group. If M has a complement C in G, then and Lemma 7.4 gives that
and hence
.
Consequently we may suppose that M does not have a complement in G. We choose such that G has type (N, K) and such that
is as large as possible. Then
by Lemma 7.5.
First suppose that . Then
and we see that
, because N is a Hall subgroup of G. Hence
. Next we let B be a batten of K that contains M. Then
, which means that for all
, B does not act faithfully on
.
Since there is some such that B acts on
avoiding L9, it follows from Definition 4.8 that B is not isomorphic to Q8. In addition
, if B is not nilpotent, by Definition 5.1. Therefore, in this case, the section B/M is a non-nilpotent batten as well. We conclude that K/M is a batten group.
Assume for a contradiction that , but that
. Then we note that
by Lemma 7.2, and we deduce that
, because M is a minimal normal subgroup of G. This forces
. Then
is not elementary abelian, because otherwise M would have a complement in this commutator and hence in G. But we are working under the hypothesis that it does not.
Now Lemmas 5.4 and 4.17(b) imply that . But now
is the unique subgroup of order 2 of
, which means that it must be centralized by K and contained in M. But this is a contradiction. Thus
and Corollary 5.6 implies that p = 2. We summarise that
and that K/M is a batten group or
and p = 2.
Let be the natural homomorphism. Then
, where
is a normal nilpotent Hall subgroup of G with modular Sylow subgroups, since sections of modular p-subgroups are modular. Moreover
is a batten group. Thus
satisfies
of Definition 6.4. We further deduce
from Lemmas 5.2 and 5.7.
For we let Q be a Sylow subgroup of K and we let
be distinct primes such that
. Then
and therefore
. Since
or
, it follows that
.
We finally let be abelian with non-cyclic Sylow subgroups and
be cyclic with
and we let
be an abelian
-invariant subgroup of
such that
and
. We set
.
Then we assume for a contradiction that every nontrivial element centralizes neither
nor
. Then Lemma 1.1 yields that
. We choose pre-images H, L and J in G of smallest possible order. Then
and
for all
, because G is soluble by Lemma 7.1. In particular we have that
.
If for some
, then r = 2. Then our assumption implies that
. It follows that
or that
. In the second case, we apply Lemmas 5.4 and 4.17. Together they show that
and thus
. This means that
centralizes
or
. But then we also have that
for some
, which is a contradiction. We deduce that
and
. In particular
is abelian, with non-cyclic Sylow subgroups, and
is an abelian L-invariant subgroup of N. Since
is cyclic, it follows from our arguments above that L is also cyclic. Moreover
, because
or
.
We now investigate the action of L on H. Since and
or
, we see that L induces power automorphisms on H. In addition Lemma 1.1 shows that
and then the fact that
yields that
. Altogether we obtain, by applying (
) to G, some
such that g centralizes
or
, where
. Since
, it follows that
and
or
. Again we use that
acts on
and
equivalently to L, because
. Then we see that
This is a contradiction. □
Lemma 7.7.
Let be of type (N, K) such that
, let
and let
. Then
has prime power order.
Proof.
We apply Lemma 7.4 and we see that has type
. Since
does not centralize N, there is a prime
such that
. It follows that
. Now
implies that
centralizes
for every
, and this shows that
. □
Lemma 7.8.
Let be of type (N, K) such that
, let
and let
. Let
be such that
does not centralize
. Then the following hold:
.
.
acts transitively on
.
If
, then there is some
such that
.
Suppose that
, that q divides
and that
. Then
if and only if
.
Proof.
We set . Then Lemma 7.7 implies that
and so
. Furthermore K acts on P avoiding L9 and then we have that
by Lemma 5.4. Next
from Lemma 2.6.
Since N is nilpotent, we conclude that . But we also have that
, whence
by Lemma 5.5. Consequently
and it follows that
, as stated in (a).
For (b) we recall that, by (a), the subgroups K and normalize
. Then
. Moreover
and Lemma 1.1 implies that
, where
and therefore
as stated in (b).
From there we deduce that P0 acts transitively on by conjugation. The second statement of (c) follows because G is soluble (Lemma 7.1), together with the fact that
is the unique subgroup of its order in the Hall subgroup K of G (Lemma 2.6). This means that every subgroup of order q of G is conjugate to
, completing (c).
For (d) and (e) we let . Lemma 7.3 provides some
such that
. Moreover, (a) implies that Kg and
normalize
, and then we summarise:
Using (b) we see that , and then we take
and
such that g = yx. Now we deduce that
, as stated in (d).
Finally, suppose that q divides and that
. Suppose first that
. Then q divides
, which provides a subgroup Q0 of order q in
and, by (d), there is some
such that
. Then
and
are subgroups of
and therefore
by Lemma 1.1. We conclude that
.
Now, conversely, suppose that . Then (d) provides some
such that
. In the paragraph above we have shown that
. We apply (c) to
, which is a group in
by Lemma 7.4, and we obtain some
such that
. We conclude that
because
. □
Lemma 7.9.
Let be of type (N, K) and let
such that K induces non-trivial power automorphisms on
.
Then for all , there is some
such that
.
In addition i = 0 if and only if there is some such that for both
we have
.
Proof.
We first remark that Lemma 7.2 gives that every subgroup of P is normal in N, and hence in G, because K normalizes every subgroup of P as well. In addition is elementary abelian by Corollary 5.6 (c).
Let . Then Lemma 7.3 provides
such that
and
.
This implies that , bearing in mind that
and
are normal subgroups of G, and therefore
Since P is elementary abelian, we see that and we deduce the first assertion.
If it is possible to choose x = y, then and in particular i = 0 in the statement of the lemma.
For the converse we suppose that i = 0, i.e. . Then
and thus we find
and
such that
. Then
. We note that x0 normalizes X, centralizes
and normalizes
, which implies that
. Similarly
. □
Lemma 7.10.
Let be of type (N, K). Suppose that X and Y are subgroups of G such that
and let B is a batten of K. Suppose that K has a normal q-complement H. Then one of the following hold:
,
, or
q = 2, K has a section isomorphic to Q8 and 4 divides
.
Proof.
Let and suppose that q divides neither
nor
. Then Lemma 7.3 gives maximal subgroups MX and MY of Q such that
and
.
If Q is cyclic, then and it follows that
. This is impossible. We conclude that Q is not cyclic and then, since K is a batten group, it follows that
. We assume for a contradiction that
. Then
and hence
, which is again a contradiction. □
Lemma 7.11.
Let be of type (N, K) such that K acts irreducibly on
for some prime
. If
are such that
, then X or Y acts irreducibly on
.
Proof.
Let be such that
and let
. We assume for a contradiction that neither X nor Y act irreducibly on
. Lemma 7.2 implies that N normalizes every subgroup of P (*). Moreover, by Lemma 7.3, there are
such that
and
and, by assumption, neither
nor
act irreducibly on
. It follows from Lemma 5.4 that
and
both induce power automorphisms on P and that
. Thus (*) yields that X and Y normalize every subgroup of P. Then also
normalizes every subgroup of P, which contradicts the irreducible action of K. □
8 The main result
Theorem 8.1.
If , then G is L9-free.
Proof.
Assume for a contradiction that the statement is false and let G be a minimal counterexample. Then there is a sublattice of L(G) isomorphic to L9, and in particular
satisfies the relations in Definition 3.1.
We let G be of type (N, K) where, among the minimal counterexamples, we choose G such that is as large as possible and we set
Then K has a normal q-complement for every and Lemma 7.10 is applicable.
We will first analyze how fits into the subgroup lattice of G.
(1) F = G, and every subgroup of N is normal in N.
Proof. The group F is a subgroup of G that is not L9-free, and Lemma 7.4 yields that . Hence the minimal choice of G implies that F = G.
Similarly, it follows from Lemma 3.4 that G is not a direct product of two non-trivial groups of coprime order. Let . Then
because N is nilpotent. If K centralizes
, then
, where the direct factors have coprime order by (
). But we just saw above that such a direct decomposition of G is not possible. Therefore
and p divides
, which divides
. We conclude from Lemma 7.2 that every subgroup of
is normal in N. This implies that every subgroup of N is a normal subgroup of N, because N is nilpotent.
Since we have chosen N as large as possible, Lemma 7.5 implies that . Let
. Then the previous equation forces
, and therefore a Sylow q-subgroup of K has order at least q2. In particular, for all non-nilpotent battens V of K, we have that
. It follows that
. Now Lemma 7.10 provides
such that
and
. Since
is characteristic in
, it follows that
is not L10-free. Moreover,
by Lemma 7.6. Since G is a minimal counterexample, we conclude that
. We recall that G is soluble, by Lemma 7.1, hence
is soluble, and then there must exist a prime
such that
. This gives a contradiction, and therefore
. □
We remark that, by (1), we may apply Lemmas 7.7 and 7.8.
(2) For all we have that
. In particular
and
.
Proof. Let , let
and set
. If H = E, then we set
and otherwise we set
. Then for all distinct
, we have that
and
.
Assume for a first contradiction that is not a normal subgroup of G. Since every subgroup of N is normal in N by (1), it follows that K does not induce power automorphism on P. Then Lemma 5.4 implies that K acts irreducibly on
. We apply Lemma 7.11 twice to find some
and some
such that X and Y act irreducibly on
. In particular
. It follows that
, which yields that
is normalized by K and hence
. This is a contradiction. We deduce that
as stated, in particular
and also
.
For the final statement in (2) we use that is not L9-free. Then the minimality of G and Lemma 7.6 give that
. □
(3) For every such that
, one of the following holds:
or K induces power automorphisms on
and
.
Moreover E = 1.
Proof. We adopt the same notation as in the previous step, which means that and
. If H = E, then
, and otherwise
. Whenever
are distinct, then
and
.
Let and
. By conjugation we may suppose that
. Then
for some Sylow q-subgroup Q0 of K by Lemmas 2.6 and 7.7 provides some
such that
is a p-group. Now Lemma 7.8 (e) states that
for all subgroups
(*).
Let X and Y be distinct elements of and assume for a contradiction that
and
are subgroups of
. Then
which contradicts (1).
For the remainder of this proof we let X and Y in be such that their intersection with P is not contained in
. We note that
by Lemma 5.5. Then it follows that
and hence
. In a similar way we observe that
.
If K acts irreducibly on , then Lemma 7.11 yields that X or Y, say X, acts irreducibly on
. It follows that
and
. Let
. Then Lemma 7.11 yields that Z or Y, say Y, acts irreducibly on
and contains
. Since
, it follows that
. By (1), this is only possible if H = D, in other words
. Then the fact that
(see (1) once more) forces E = 1.
The previous paragraph also gives that , by Lemma 5.4, in the case where K acts irreducibly on
.
Otherwise Lemma 5.4 gives that K and hence induce power automorphisms on P. Together with (1) this means that every subgroup of P is normal in G. Now, if
are distinct, then
and it follows that
. We deduce that
for all
. In particular
for all
. This implies all our claims about D, because
by Lemma 5.4 and
.
If, still in the power automorphism case, we have that H = E, then we recall that by (1). Therefore
This implies that . But in this case we may interchange T by S in
. Since
, we arrive at the contradiction that
(by (1)). Thus, we have that
in this case as well. Again it follows that E = 1. □
The remainder of the proof is dedicated to constructing a subgroup of G that violates Property ().
(4) If , then
does not centralize
. In particular
.
Proof. We assume for a contradiction that centralizes
for some
.
It follows from Lemma 7.3 that, for every subgroup X of , there is some
such that
. Since
, we further have THAT
for all
and
(*).
Assume that . Then for both
, we see that
and therefore
. But this contradicts (3).
Thus and then (1), together with our assumption at the beginning of the proof, imply that every subgroup of
is normal in A. Suppose that
are distinct. We recall that
, and then it follows that
. Since
, we know more:
and
.
If K induces power automorphisms on P and if , then
and
are normal subgroups of G by (1). In addition there is some
such that
and then
by (*). We apply Lemma 7.9 to see that
. Now it follows that
, which is a contradiction.
We conclude that K does not induce power automorphisms on P. Then Lemma 5.4 gives that K acts irreducibly on . Since
by (2), we see that either
or that
.
In the first case and
. This implies, together with Lemma 1.1, that
has a subgroup isomorphic to
. We apply Lemma 5.4, in combination with Part (b) of Lemma 4.17, and we deduce that
is cyclic of order 2 and that, therefore, K centralizes it. This is impossible.
It follows that the second case holds, i.e. . Then p = 2 by Corollary 5.6 (a). If
is not elementary abelian, then Part (b) of Lemmas 4.17 and 5.4 give that
and
, where I is a subgroup of P of order at most 2. We note that
and
, and then we conclude that
and
are cyclic of order 2. Consequently
. Now
and therefore
. We arrive at a contradiction:
, because
is not elementary abelian.
So we finally have that is elementary abelian. Then Lemma 4.17(b) gives that
is cyclic and Lemma 1.1 shows that
.
Together with the fact that , we obtain that
is cyclic of order 2. It follows that
and
have order 2 and hence
is elementary abelian of order 4. Moreover
is cyclic of order 2, and therefore it equals one of the subgroups
or
. The last case is not possible because
. By symmetry between S and T in the lattice we may suppose that
. Recall that K acts irreducibly on
while
centralizes P. In particular (1) implies that A does not act irreducibly on
. Thus Lemma 7.11, together with the fact that
, gives that U acts irreducibly on
. We also know that
, and this implies that
. Then Lemma 1.2 yields that
. But we recall that
is cyclic, and its unique involution is contained in D. Then the fact that
implies that
. Finally, we see that
, which gives a contradiction. □
By conjugation and by Lemma 7.3 we may suppose that .
We set and we let L1 be a Hall π-subgroup of
.
Then we let
(5) If and
, then
. Moreover
.
Proof. First suppose that L1 centralizes for some
. Then (4) implies that
and then there is a non-nilpotent batten V of K such that
and
. Now
is a prime and
by Lemma 7.7. In addition we see, from Definition 4.16, that Q does not induce power automorphisms on P and that
. If q divides
, then (3) implies that
, as stated.
Now we suppose that q does not divide . Let
be such that V = QR. Up to conjugation we may suppose that
is a Sylow subgroup of L1. Then R does not centralize P by Definition 5.1 and hence
. It follows that
, from the definition of L1. Since
is nilpotent by Lemma 2.5, we deduce that A has a normal q-complement. Moreover
, by Lemma 7.4, whence we may apply Lemma 7.10 to A. Then we see that S and T have orders divisible by q, because
. Let
be such that
and
. The irreducible action of Q on P, together with the fact that
, implies that
or
. Without loss
. Then
by Lemma 7.8 (e). Assume that A normalizes Qt. Then
, which implies that
by Lemma 4.1.1 (b) of [Citation7]. Then the irreducible action of Qt on P forces t = s, contradicting the fact that
.
Thus A does not normalize Qt and , which yields that
by Part (a) of Lemma 7.8. It follows that
and therefore
, because
by (2) and K acts irreducibly on P.
We turn to the second statement and assume for a contradiction that . Then
, contrary to (4). Hence if
, then we can draw two conclusions: First
and the statement we just proved gives that
. Second, there must be a prime in
dividing
. By definition of
, such a prime is
for some non-nilpotent batten V of K. We choose such a non-nilpotent batten V and set
. Then there are some prime r and an r-subgroup
such that QR = V, and
because of the structure of non-nilpotent battens (Definition 2.1). In particular
by assumption and Lemma 7.10 yields that B and U contain a conjugate of R. We recall that
by the first consequence of our assumption and because of the structure of the lattice L9. Then Lemma 7.8(c) gives that
. Thus
, which is a contradiction. This proves that
.
If , then for all
, all
and all q-subgroups Q of L1, we have that
. Then
by definition of L1, and
because
. But then
, contrary to (1). □
Next we set and we prove that H is a candidate for the desired properties in (
).
(6) The non-trivial Sylow subgroups of H are elementary abelian (in particular N is abelian), but not cyclic, and .
Proof. By definition is nilpotent. If, for all
, the group K does not act irreducibly on
, then Lemma 5.4 gives that
. In particular
is not cyclic. Moreover, Corollary 5.6 yields that
is elementary abelian. Therefore, our claim is satisfied in this case.
Let us assume for a contradiction that there is some such that K acts irreducibly on the group
. We set
and we choose
of order
such that
, by the definition of σ.
Case 1: .
Then Lemma 7.8(c) implies that for every
. We recall that
, and then it follows first that
and then that
, by Lemma 7.10. Here we use that
by (1). Thus we find some
such that
, because
. We apply (3) to observe that
and then D contains every conjugate of Q by Lemma 7.8(c). Recall that
, which implies that
by Lemma 7.10. Again we use that
. But this contradicts the fact that
by (3).
Case 2: .
Then , in particular
, and
by (2). This means that K stabilizes the subgroup
of
, while acting irreducibly. This forces
, and together with coprime action (Lemma 1.1) we see that K centralizes
.
By Lemma 7.3 we know that , with type
. Additionally, since
, the group K has a normal q-complement. Then
also has a normal q-complement. We may apply Lemma 7.10 to
. It yields that at least two of the groups D, T, S have a subgroup of order
. As
by (2), there is some
such that
and
. Then Part (e) of Lemma 7.8 implies that
by Lemma 5.5. In the present case we have that
, and then Lemma 1.2 gives that A does not act irreducibly on
. Thus
induces power automorphism on P by Lemma 5.4, and these automorphisms are not trivial because
. Corollary 5.6 (c) implies that
is elementary abelian and in particular that
. Hence there is some
such that
, by Lemma 7.8(b). In addition we see from Lemma 7.11 that B and C act irreducibly on P. If it was true that
, then it would follow that
, which is a contradiction. We conclude that
and hence
because of the irreducible action of B on P, and similarly
.
Then Lemma 7.8(b) provides such that
and
. If
, then
, which is false.
Consequently , and we can use Lemma 7.8(a), Dedekind’s modular law and Lemma 1.4(c). Together this shows that
because N is nilpotent and because
.
We recall that is abelian, hence it is contained in Z(N), and then it follows that D centralizes
. Here we also use Lemma 5.4, i.e. that D centralizes P. We conclude that
for all
. Again Lemma 7.8(b) provides
such that
and
. Let
. Then
, which implies that
. But now
, by (3), which gives a final contradiction. □
(7) For all we have that
.
Proof. We set for some
. Then there is some
of order
such that
, by the definition of σ. Then
and we see that K as well as
have a normal q-complement. Since
by Lemma 7.4, we may apply Lemma 7.10. We notice that
, and then it follows that
or that
. In the first case (3) implies our assertion. In the second case we use Lemma 7.8(c). It gives that
and
for some
. We deduce that
by Lemma 4.1.1 of [Citation7]. In conclusion, our assertion is true or
centralizes Q. But in the second case, we see that
, and this gives a contradiction. □
(8) For all we have that q divides
, and
.
Furthermore, if and
, then
, but
.
Proof. Suppose that has order q. Then
is a p-group by Lemma 7.7, for some prime
. We set
. Then for every
there is some
such that
, by Lemma 7.8(b). Using (6) we see that K induces power automorphisms on P. Thus
is elementary abelian by Corollary 5.6(c), and then Lemma 7.9 is applicable.
Assume for a contradiction that . Then, for all
, we have that
Hence Lemma 7.9 yields that and, for all
, we see that
. In particular, we have that
. Therefore, the fact that
implies that
. This contradicts (7).
We conclude that and hence Lemma 7.9 provides some element
such that, for both
, it is true that
.
Assume for a contradiction that . Then Part (e) of Lemma 7.8 yields that
. This is impossible. Thus
and we deduce that
for all
. Here we use Lemma 7.10 and the fact that
. Next we apply Lemma 7.9, which gives that
. Then the same lemma yields, for both combinations of
, that p divides
It follows that and then that
. Finally (6) and Lemma 7.9 yield that
. We conclude that
. □
(9) For every there is some
such that a Sylow r-subgroup of U does not centralize
.
Proof. We set . Since
, there is some
such that
. In addition (6) implies that P is elementary abelian, and then
because N is nilpotent. By Lemma 7.8(b) and (8) there is some
such that
.
Assume for a contradiction that U centralizes P. Then by Lemma 7.8(d) and hence
implies that
. With (7) we obtain the contradiction that
.
It follows that U does not centralize P. We recall that and then we obtain a prime
such that a Sylow r-subgroup of U does not centralize P. Since K induces power automorphisms on P by (6), Definition 5.1 gives, for every non-nilpotent batten V of K, that
centralizes P. This implies that
. □
We are now able to define L and J. Let be such that
.
We set and we let L2 be a Hall
-subgroup of
. Then
and
is a subgroup of K. In addition (5) and (9) show that
.
Next we set and
. Then
by Lemma 7.7, because
, and hence
. Finally, we note that J is L-invariant by construction.
(10) , i.e.
and
are coprime.
Proof. We assume for a contradiction that the prime p divides and
. Then by definition there are
and
and subgroups
and
such that the following hold:
and
.
In particular (6) shows that K induces power automorphisms on P. It follows from Corollary 5.6 (c) and Lemma 5.4 that , and then (8) shows that
. Moreover
by (7). We apply Lemma 7.9 to obtain some
such that
This implies that , and we obtain that i = 1 and j = 0.
In particular we see that . Then Lemma 7.9 and the fact that
give some element
such that
, where
. Assume for a contradiction that
. Then 7.8(d) forces
. This is impossible, hence
and we apply Lemma 7.10 to
. Then it follows that
.
Moreover by (8). Since
, this is not possible, hence Lemma 7.9 and (8) give that
. On the other hand we have that
, which is also impossible because
. Now Lemma 7.9 implies that
. In particular we see that
, and this contradicts (8). □
We summarize:
The definitions of σ and H imply that , and (6) shows that the non-trivial Sylow subgroups of H are not cyclic. In addition
induces power automorphisms on H.
From (10) we deduce that . A Hall
-subgroup of K is nilpotent by Lemma 2.7, which means that L is nilpotent. In particular we see that
. Let
and
. If
, then
, and if
, then
by (9). These arguments show that
(*).
We assume for a contradiction that L is not cyclic. Then, since L is nilpotent and a batten group by Lemma 2.7, we deduce that . Definition 4.8 implies that, for all
, the group
does not induce non-trivial power automorphisms on
. In particular
centralizes H, which contradicts (*). Thus L is cyclic.
Furthermore, we already saw that J is L-invariant and that . Then (10) gives that
, and since N is nilpotent, this forces
.
Assume for a contradiction that J is not abelian. Then Lemma 4.17(b) and Lemma 5.4 show that and therefore
. We let
be such that a Sylow r-subgroup R of K acts faithfully on
. Then we must have that
because
. Moreover Lemma 7.8(a) yields that R centralizes
. This contradicts (*).
Hence J is abelian. For every and every subgroup
of order q, the definition of σ and Lemma 7.7 provide some
such that
. Then we see, using (6) and Corollary 5.6, that
. Furthermore (10) yields that Q centralizes
, and then it follows that
. Since
is abelian, we conclude that
.
The previous argument also yields that . Let
and suppose that
has order r. We recall the definition of ρ and apply Lemma 7.7: Then we see that
. Thus
and it follows that
.
Since , we can use Property
, which gives some
that centralizes
or
. Let
be such that
. We may suppose that g has prime order p. Then
and hence there is a subgroup
of prime order such that
. Lemma 5.5 gives that
or that Li centralizes P.
In the first case Corollary 5.6 yields that p = 2, and then K does not induce power automorphisms on P. Using (6) we deduce that and
. Since J is abelian, we obtain a contradiction in this case.
It follows that the second case above holds, i.e. . This means that i = 1 if
and i = 2 if
. In addition we see, from (9), that
. We conclude that
and
. In particular
. Since
, Lemma 7.10 implies that B and U contain a conjugate of Q. In addition (5) shows that
and therefore Lemma 7.8(c) gives that
. Finally, we obtain a contradiction, because
by (3). This concludes the proof. □
Main Theorem. A finite group is in if and only if it is L9-free.
Proof. Let G be a finite group. If G is L9-free, then Theorem 6.5 shows that .
Conversely, if , then G is L9-free by Theorem 8.1. □
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