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Research Articles

Projective characters of metacyclic p-groups

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Pages 2736-2747 | Received 02 Jun 2022, Accepted 18 Jan 2023, Published online: 05 Feb 2023

Abstract

Let ω be a 2-cocycle of a metacyclic p-group G representing a non-trivial element of the Schur multiplier M(G). Then the number of ω-regular conjugacy classes of G, the subgroup consisting of the ω-regular elements in the center of G, the degree of each irreducible ω-character of G and a representation group H of G with M(H) trivial are all determined. Finally, for ω constructed from H, the projective character table of G corresponding to ω is found in the case that G is of positive type.

Communicated by Mark Lewis

2020 Mathematics Subject Classification:

1 Introduction

All groups considered in this paper will be finite and such a group G is metacyclic if there exists a cyclic normal subgroup N of G such that G/N is cyclic. Equivalently G is metacyclic if and only if there exists a short exact sequence 1CmGCs1for some m,sN, where Cn denotes a cyclic group of order n (under the operation of multiplication).

Let G be a group and let ω be a 2-cocycle of G over the field of complex numbers C, so that ω(1,x)=1=ω(x,1) and ω(x,yz)ω(y,z)=ω(x,y)ω(xy,z) for all x, y, zG.

The set of all such 2-cocycles of G form a group Z2(G,C*) under multiplication. Two 2-cocycles ω and ψ are cohomologous if there exists a function δ:GC* such that ψ=t(δ)ω, where t(δ)ω(x,y)=δ(x)δ(y)δ(xy)ω(x,y)for all x,yG. This defines an equivalence relation on Z2(G,C*) and the cohomology classes [ω] form a finite abelian group, called the Schur multiplier M(G).

Definition 1.1.

Let ω be a 2-cocycle of a group G. Then xG is ω-regular if ω(x,g)=ω(g,x) for all gCG(x).

Let Gω be the set of ω-regular elements of G. It is easy to check that if ψ[ω], then Gψ=Gω. Moreover, it is well-known that any conjugate of an ω-regular element is also ω-regular (see [7, Lemma 2.6.1]), so that we may refer to the ω-regular conjugacy classes of G, meaning the conjugacy classes of G that consist of ω-regular elements. However, if x,yGω then it does not follow in general that xyGω.

We next define Zω(G)=GωZ(G), where Z(G) denotes the center of G. This set was studied in [12, Section 3] and in particular Zω(G) is a subgroup of G. A further refinement of this concept is the epicenter Z*(G) of G, which is defined by Z*(G)=[ω]M(G)Zω(G).

The group G is said to be unicentral if Z*(G)=Z(G), and to give a flavor of the results to come concerning metacyclic groups, we have [3, Proposition IV.4.21]:

Lemma 1.2.

Let G be a metacyclic group. Then M(G) is trivial if and only if G is unicentral.

To progress further we next introduce projective representations of G, using the general linear group GL(n,C) of invertible n × n matrices over C.

Definition 1.3.

Let ω be a 2-cocycle of a group G. Then P:GGL(n,C) is an ω-representation of G if P(1)=In and P(x)P(y)=ω(x,y)P(xy)

for all x,yG.

An ω-representation is also called a projective representation of G with 2-cocycle ω and its trace function is its ω-character.

The ω-representation P is said to be irreducible if there does not exist an invertible matrix M such that MP(g)M1=P1(g)P2(g) for all gG, where P1 and P2 are also ω-representations of G. The set of irreducible ω-characters is denoted by Proj(G,ω) (or Irr(G) in the case ω is trivial) and the cardinality of this set is equal to the number of ω-regular classes of G (see [8, Theorem 1.3.6]). It is also the case that xGω if and only if ξ(x)=0 for all ξProj(G,ω) or equivalently that xGω if and only if there exists ξProj(G,ω) such that ξ(x)=0 (see [8, Proposition 1.6.4]).

Given a 2-cocycle ψ of G it is well-known (see [4, Lemma 4.5]) that one may choose ω[ψ] such that the order of ω equals the order of [ψ] and also so that ω is a class-preserving 2-cocycle of G. The latter condition means that the ω-characters of G are constant on the conjugacy classes of G. For such a choice satisfying both conditions the normal “row” orthogonality relations hold for the irreducible ω-characters of G and the normal “column” relations hold when restricted to Gω. Here row and column refer to the rows and columns of the square projective character table of G, in which any columns corresponding to non-ω-regular conjugacy classes are omitted. Generally we will always assume that we are starting with a 2-cocycle ω with the two properties given above.

One of the ways of finding representative projective character tables of a group G is to find the ordinary character table of a representation group H for G.

Definition 1.4.

Let G be a group. Then a group H is called a representation or covering group for G if there exists AZ(H)H such that AM(G) and H/AG.

Details of the following basic concepts may be found in [Citation7] or [Citation4]. A representative class-preserving 2-cocycle ω for each [ω]M(G) can be constructed from a representation group H of G using each linear character λIrr(A) and the order of ω is also equal to that of [ω]. This is achieved using the transgression isomorphism tra:Irr(A)M(G). In fact tra(λ) produces a specific ω for a fixed choice of conjugacy-preserving transversal of A in H but the cohomology class of ω is independent of the choice made. Moreover the elements of Proj(G,ω) are linearized by Irr(H|λ), where the latter set denotes the set of χIrr(H) such that λ is a constituent of χA. It is also the case that Z*(G)Z(H)/A from [12, Theorem 3.5].

Given ω constructed from λIrr(A) as above, we set K=kerλ and H¯=H/K. Then H¯ is called an ω-covering group for G and ω in the sense that ω may be constructed from (with the obvious notation) λ¯Irr(A¯) and Proj(G,ω) are linearized by Irr(H¯|λ¯). Again it is the case that Zω(G)Z(H¯)/A¯, this time tra defines an isomorphism from Irr(A¯) to [ω].

In Section 2 we will consider groups G for which Gω=Zω(G) and show that the elements of Proj(G,ω) are particularly easy to describe in this special case, which in particular covers all abelian groups. Another observation in this situation is that any 2-cocycle ω of G is automatically class-preserving, since the only ω-regular conjugacy classes of G consist of central elements. In Section 3 we consider metacyclic p-groups and in particular the division of these groups into positive and negative type when p=2. In Sections 4 and 5 the Schur multiplier, a representation group, the ω-regular elements and the corresponding projective character tables for all [ω][1] are found for any metacyclic p-group of positive type. In Section 6 similar calculations are performed for a metacyclic 2-group G of negative type, this time the Schur multiplier, a representation group, the number of ω-regular conjugacy classes, Zω(G) and the degree of each element of Proj(G,ω) are all described.

2 Groups in which the ω-regular elements are all central

Let G be a group and let N be a normal subgroup of G. Then the inflation homomorphism, inf:M(G/N)M(G) is defined by inf([ω¯])=[ω], where ω(x,y)=ω¯(xN,yN) for all x,yG. For notational convenience we will also define ω=inf(ω¯) in this context.

Lemma 2.1.

Let ω be a 2-cocycle of a group G, let U=Zω(G) and let xZ(G). Then

  1. [ωU] is trivial;

  2. [ω] is in the image of inf:M(G/U)M(G);

  3. if ξProj(G,ω), then ξU=ξ(1)μ for some μProj(U,ωU);

  4. xU if and only if ξ(x)=0 for all ξProj(G,ω);

  5. xU if and only if ξ(x)=0 for some ξProj(G,ω);

  6. if ξProj(G,ω), then there exists [ψ¯]M(G/U) with [ω]=inf([ψ¯]) such that ξ=ζκ for ζ¯Proj(G,ψ¯) and some κProj(G,ωψ1), where if ξU=ξ(1)μ then κU=μ.

Proof.

Parts (b), (c), (d), and (e) are all proved in [12, Theorems 4.4 and 3.4]. Part (a) can be proved in a variety of ways, for example o([ω]) divides ξ(1) for all ξProj(G,ω), and μ(1)=1 in part (c). Part (f) is just an application of Clifford’s Theorem for projective characters in this special case (see [8, Theorem 2.2.1]). □

If ω is a 2-cocycle of a group G, then G is said to be of ω-central type if |Proj(G,ω)|=1. This means that the corresponding projective character table of G has just one single entry of |G|. We also note in passing that if [ω] is trivial, then Proj(G,ω)={δχ:χIrr(G)} for δProj(G,ω) with δ(1)=1.

Proposition 2.2.

Let ω be a 2-cocycle of a group G such that Gω=Zω(G). Then, with U=Zω(G),

  1. ξ(1)2=|G:U| for all ξProj(G,ω);

  2. [ω]=inf([ψ¯]) for some [ψ¯]M(G/U) and G/U is of ψ¯-central type for all such ψ¯.

Proof.

Using the notation of Lemma 2.1 (f), κ(1)=1 and so ζ¯(gU)=0 for all gU and hence G/U is of ψ¯-central type. Finally ξ(1)2=ζ¯(U)2=|G:U|.

We may vary Proposition 2.2 by replacing ω by inf(ψ¯) as shown in Lemma 2.1 (f), so that Proj(U,ωU)=Irr(U). The price for doing this is that the order of the original 2-cocycle ω now divides that of ψ¯ and ψ. This change of 2-cocycle clearly doesn’t affect part (a) of Proposition 2.2, but in general in part (f) of Lemma 2.1 the 2-cocycle ψ that arises depends on the choice of ξ and in particular ωψ1 is not always trivial.

Our next result demonstrates that the variation considered above can have other drawbacks.

Lemma 2.3.

Let ω be a 2-cocycle of a group G. Let U=Zω(G) and V=Zωn(G) for nN, and let infU,infV denote the inflation homomorphism from M(G/U) and M(G/V) respectively into M(G). Then

  1. GωGωn;

  2. there exists [ψ¯]M(G/U) and [ϕ¯]M(G/V) such that

  3. infU([ψ¯])=[ω] and infV([ϕ¯])=infU([ψ¯]n).

Proof.

For (a), it is obvious that if gG is ω-regular, then it is also ωn-regular.

For (b), from Lemma 2.1(b) there exists [ψ¯]M(G/U) such that infU([ψ¯])=[ω]. Now [ω]n=infUn([ψ¯])=infU([ψ¯]n).

Next every element of V is ωn-regular and so every element of V is ψn-regular. It follows from Lemma 2.1(d) that every element of V/U is ψ¯n-regular Thus from Lemma 2.1(b) there exists [ϕ¯]M((G/U)/(V/U)) with inf([ϕ¯])=[ψ¯]n. Finally, identifying the group (G/U)/(V/U) with G/V yields the desired result. □

In the context of Lemma 2.3 we have shown that we may replace ω by infU(ψ¯), but it may not be the case that we can also replace ωn by infV(ψ¯n).

3 Metacyclic p-groups

Metacyclic groups G have presentations of the form G=a,b:am=1,bs=at,bab1=ar,where m, r, s, and t are positive integers satisfying rs1(modm) and m|t(r1). Conversely if the positive integers m,s,r,t satisfy these two conditions, then the given presentation defines a metacyclic group of order ms. If G has the above presentation, then we may replace a by ak, where k=t/(m,t), which means that we can, and will, assume that t|m. Now Beyl, (see [3, pp. 194–195]), defines t0=m/(m,r1) and λ=t/t0, so that G has parameters m, r, s and λ.

It is convenient to deal with cyclic groups separately at this juncture. For if G is any cyclic group then M(G) is trivial (see [6, Proposition 2.1.1]) and the elements of Irr(G) are well-known. To save repetition henceforth the generators of a non-cyclic metacyclic group G will normally be denoted by a and b.

In [9, Section 1] a standard parametric presentation is given of non-cyclic metacyclic p-groups using the following notation: Gp(α,β,ϵ,γ,±)=a,b:apα=1,bpβ=apαϵ,bab1=apαγ±1,with details as follows for firstly abelian and secondly non-abelian groups.

Lemma 3.1.

Let G be a non-cyclic abelian metacyclic p-group. Then there exist α,βN and ϵ a non-negative integer with ϵ<α such that for an odd prime number p, GGp(α,β,ϵ,0,+); whereas for p=2,GG2(α,β,ϵ,0,+) for α=1 or GG2(1,β,0,0,).

We note that, using the exponent of G and the classification of abelian groups, we have for G in Lemma 3.1 and αβ+ϵ that G=Gp(α,β,ϵ,0,+)Gp(α,β,0,0,+) and for α<β+ϵ that G=Gp(α,β,ϵ,0,+)Gp(β+ϵ,αϵ,0,0,+) for all prime numbers p. So in particular for p odd, a non-cyclic abelian metacyclic p-group is isomorphic to exactly one group of the form Gp(α,β,0,0,+) for αβ1. The next result is now obvious from these remarks.

Lemma 3.2.

Let α,βN and let ϵ be a non-negative integer with ϵ<α. Then

  1. for p odd, Gp(α,β,ϵ,0,+) is a non-cyclic abelian metacyclic p-group of order pα+β;

  2. for p=2,G2(α,β,ϵ,0,+) is a non-cyclic abelian metacyclic 2-group of order 2α+β for α=1 and G2(1,β,0,0,) is a non-cyclic abelian metacyclic 2-group of order 21+β.

We now turn to non-abelian groups where the next result is [1, Theorem 1.2].

Theorem 3.3.

Let G be a non-abelian metacyclic p-group. Then there exist non-negative integers α,β, γ, and ϵ with α,β>0,γmin{α1,β} and γ+ϵα such that for an odd prime number p, GGp(α,β,ϵ,γ,+).

If p=2, then in addition αγ>1 and GG2(α,β,ϵ,γ,+) or G2(α,β,ϵ,γ,),

where in the second case ϵ1. Moreover, if GGp(α,β,ϵ,γ,+) then γ>0 for all prime numbers p.

A non-cyclic metacyclic p-group G is of positive type if GGp(α,β,ϵ,γ,+) for any prime number p and of negative type if p = 2 and GG2(α,β,ϵ,γ,), for some parameters satisfying the conditions in Lemma 3.1 or Theorem 3.3. It is possible in the case p = 2 for G to be both of positive and negative type for different parameters and this is discussed below.

We show that if the conditions on the parameters α,β, ϵ and γ in Theorem 3.3 are satisfied then Gp(α,β,ϵ,γ,±) defines a metacyclic p-group. The proof utilizes the p-adic valuation vp(n) of an integer.

Lemma 3.4.

Let α,β, γ and ϵ be non-negative integers with α,β>0,γmin{α1,β} and γ+ϵα. Then

  1. Gp(α,β,ϵ,γ,+) is a non-abelian metacyclic group of order pα+β if γ>0 and if in addition for p=2,αγ>1;

  2. G2(α,β,ϵ,γ,) is a non-abelian metacyclic group of order 2α+β if αγ>1 and ϵ1.

Proof.

We first note that αϵ0, so pαϵ|pα; also if p=2, then αγ>1 so 2αγ1>1. Now we show that the conditions ensure that (pαγ±1)pβ1(modpα) and pα|pαϵ((pαγ+1)1) in case (a) or 2α|2αϵ((2αγ1)1) in case (b).

First we observe that ϵ+γα, so in case (a) pαϵ((pαγ+1)1)=p2α(γ+ϵ) and this is divisible by pα. On the other hand in case (b), ϵ{0,1} and ϵ+γα, so 2αϵ(2αγ2)=22α(ϵ+γ)2αϵ+1=2α(2α(ϵ+γ)21ϵ) and this is divisible by 2α. Now βγ0 and for p=2,αγ>1, so vp((pαγ±1)pβ1)=αγ+βα from [1, Lemma 2.3]. □

Now any metacyclic 2-group of nilpotency class at least three is not of both positive and negative type from [1, Lemma 3.1]. We observe that, up to isomorphism, G2(1,1,0,0,)C2×C2 is the only group in Lemma 3.1 that is only of negative type, whereas G2(1,β,0,0,)C2×C2β for β>1 are the only groups in that lemma that are both of positive and negative type. It also follows from [1, Corollary 2.6] that, up to isomorphism, G2(2,β,0,0,) and G2(2,β,1,0,) are the only groups of nilpotency class two in Theorem 3.3 that are of negative type. Beuerle [1, Section 1] assigns a unique set of parameters for all non-abelian metacyclic p-groups up to isomorphism, however we only need to consider 2-groups of nilpotency class one or two.

Lemma 3.5.

A non-cyclic abelian metacyclic 2-group is isomorphic to exactly one group in the following list of such groups: G2(α,β,0,0,+) for αβ1 and (α,β)(1,1) or G2(1,1,0,0,).

We next consider metacyclic 2-groups of nilpotency class two, where the results are a slight variation of [1, Theorem 1.1].

Theorem 3.6.

Let G be a metacyclic 2-group of nilpotency class two. Then G is isomorphic to exactly one group on the following list:

  1. G2(α,β,0,γ,+), where α,β,γN with α2γ,βγ and (α,γ)=(2,1).

  2. G2(2,β,0,0,) for βN.

  3. G2(2,1,1,0,), the quaternion group of order 8.

Notice that this result implies that the groups G2(2,β,1,0,) for β>1 are of both positive and negative type.

Lemma 3.7.

The groups G2(2,β,0,0,) for βN and G2(2,1,1,0,) are not of positive type.

Proof.

Let G=G2(2,1,0,0,) or G2(2,1,1,0,). Then Z(G)=a2 and so if N is a cyclic normal subgroup of G with G/N cyclic, then |N|=4.

Now suppose G=G2(2,β,0,0,) with β2. Then G=a2 from [1, Proposition 2.5] and G/GC2×C2β. Suppose N is a cyclic normal subgroup of G with 8||N| and G/N cyclic. Let aibjN of order 8. Then (aibj)4G=(a4iG)(b4jG)=b4jG=G, but (aibj)2G=(a2iG)(b2jG)=b2jGG and we conclude that j=β2. Now from [1, Lemma 2.4], (aib2β2)4=aik, where k=j=0332β2j, but v2(k)2 from [1, Lemma 2.3]. Thus (aib2β2)4=1 and hence no such N exists. □

In general we will consider metacyclic p-groups whose parameters just satisfy the conditions in Lemma 3.1 and Theorem 3.3. The degrees of freedom that these parameters afford are discussed in detail in [1, Section 3] for groups of nilpotency class at least three.

Finally in this section we provide a conversion between Beuerle’s notation in [Citation1] and that of Beyl: m=pα,s=pβ,t=pαϵ and r=pαγ±1. For groups of positive type, this leads to t0=pα/(pα,pαγ)=pγ and λ=t/t0=pα(γ+ϵ). For groups of negative type we know that p = 2 and ϵ{0,1}. Therefore t0=2α/(2α,2αγ2)=2α1 and λ=21ϵ.

4 Multipliers and representation groups of metacyclic p-groups of positive type

To save repetition we will always assume in this section and Sections 5 and 6 that any given metacyclic p-group is non-cyclic. We also shorten the notation Gp(α,β,ϵ,γ,+) to Gp(α,β,ϵ,γ), since we will only be working with groups of positive type in this and the next section.

We first present a known result concerning the Schur multiplier of metacyclic groups (see [6, Theorem 2.11.3]).

Theorem 4.1.

Let G be a metacyclic group, say G=a,b:am=1,bs=at,bab1=ar,where the positive integers m, r, s and t satisfy rs1(modm),m|t(r1) and t|m.

Then M(G)Cn, where n=(r1,m)lm and l=(1+r++rs1,t).

Using the conversion formula at the end of Section 3 we now obtain the following result for metacyclic p-groups of positive type (the calculation for groups of negative type will be postponed until Section 6).

Corollary 4.2.

Let GGp(α,β,ϵ,γ). Then M(G){Cpβγ,if αβ+ϵ;Cpα(ϵ+γ),if α<β+ϵ.

Proof.

Applying the conversion described in Section 3 to Theorem 4.1 we obtain that M(G)Cn where n=(r1,m)lm=(pαγ,pα)lpα=lpγ,with l=(i=0pβ1(pαγ+1)i,pαϵ). Now vp(i=0pβ1(pαγ+1)i)=β, from [1, Lemma 2.3] and so l=pmin{β,αϵ}={pβ,if αβ+ϵ;pαϵ,if α<β+ϵ.

Beyl [Citation2] called a group G with M(G) trivial a Schur group and we adopt this terminology. So as observed in Section 3 all cyclic groups are Schur groups. Using Corollary 4.2 we can classify all (non-cyclic) metacyclic p-groups of positive type that are Schur groups. This result confirms part of the classification of all non-cyclic metacyclic p-groups that are Schur groups given in [2, Section 5].

Corollary 4.3.

Let G=Gp(α,β,ϵ,γ). Then G is a Schur group if γ=β for αβ+ϵ, or ϵ=αγ for α<β+ϵ. The non-isomorphic metacyclic p-groups of positive type that are Schur groups are Gp(α,β,0,β), and Gp(α,β,αγ,γ) for γ<β.

Proof.

The first part is immediate from Corollary 4.2 once we note that if αβ+ϵ, then GGp(α,β,0,γ) from [1, Proposition 3.6]. For the second part if γ=β and ϵ=0, then αβ+ϵ. On the other hand if ϵ=αγ, then β+ϵ=α+(βγ)α with equality if and only if β=γ. In the latter case we obtain Gp(α,β,αβ,β)Gp(α,β,0,β) again from [1, Proposition 3.6]. Finally the groups Gp(α,β,0,β), and Gp(α,β,αγ,γ) for γ<β are all non-isomorphic from the remark before Lemma 3.2 and then also Lemma 3.5, Theorem 3.6 and [1, Propositions 3.2, 3.5 and 3.6]. □

Next we turn our attention to representation groups, for which the main result we use is [3, Theorem 2.10].

Theorem 4.4.

Every metacyclic group has a metacyclic group which is a Schur group among its representation groups.

In the context of Theorem 4.4 it should be noted that an example of a metacyclic group with two non-isomorphic representation groups that are also Schur groups is given in [3, Example IV.2.12]. We will make a definitive choice for a representation group that is also a Schur group for each metacyclic p-group of positive type with a given set of parameters.

The following result, [1, Proposition 2.5], will subsequently prove very useful.

Lemma 4.5.

Let G=Gp(α,β,ϵ,γ). Then

  1. Z(G)=apγ,bpγ and |Z(G)|=pα+β2γ,

  2. G=apαγ.

Theorem 4.6.

Let G=Gp(α,β,ϵ,γ) with αβ+ϵ. Define the sequence

  1. H(i)=Gp(β+i,β,ϵ,γ+βα+i) for i=0,,αγ if β>αγ;

  2. H(i)=Gp(αγ+i,β,ϵ,i) for i=0,,β if βαγ.

Then H(i) is a metacyclic p-group and |M(H(i))|=pαγi in case (a) and pβi in case (b). Suppose [ω]M(H(i)) of order pj. Then H(i+j) is an ω-covering group for H(i), in particular the last group in each sequence is a Schur group that is a representation group for all the groups in the sequence.

Proof.

In case (a), β+iβ0. By definition γ+βα+i>0, also β(γ+βα+i)=αγi0 and (β+i)(γ+βα+i+ϵ)=α(γ+ϵ). Thus, H(i) is a metacyclic p-group from Lemma 3.2, |M(H(i)|=pαγi from Corollary 4.2 and in particular H(αγ) is a Schur group.

Let H(i+j)=x,y. Then by Lemma 4.5, Z(H(i+j))=xpγ+βα+i+j,ypγ+βα+i+j and H(i+j)=xp(β+i+j)(γ+βα+i+j)=xpαγ. Therefore, Z(H(i+j))H(i+j)=xpk, where k=max{γ+βα+i+j,αγ}max{β,αγ}=β. Thus, Ai=xpβ+iZ(H(i+j))H(i+j). Next set a=xxpβ+i and b=yxpβ+i and observe that H(i+j)/AiH(i). Finally, |Ai|=p(β+i+j)(β+i)=pj=o([ω]) and if we let trai:Irr(Ai)M(H(i)) be the transgression homomorphism, then [ω]Im(trai) since M(H(i)) is cyclic.

In case (b), αγ+iβ0 and (αγ+i)(i+ϵ)=α(γ+ϵ). Thus, H(i) is a metacyclic p-group from Section 3, |M(H(i)|=pβi from Corollary 4.2 and in particular H(β) is a Schur group.

Using the same notation as in case (a), we have by Lemma 4.5 that Z(H(i+j))=xpi+j,ypi+j and H(i+j)=xpαγ. Therefore, Z(H(i+j))H(i+j)=xpαγ. Thus Ai=xpαγ+iZ(H(i+j))H(i). Next set a=xxpαγ+i and b=yxpαγ+i and observe that H(i+j)/AiH(i). Finally, |Ai|=p(αγ+i+j)(αγ+i)=pj=o([ωj]) and the subsequent argument is the same as in case (a). □

Notice in Theorem 4.6 that G=Gp(α,β,ϵ,γ)Gp(α,β,0,γ) from [1, Proposition 3.6], so that ϵ can be taken to be 0 in the statement and proof of the result. The group G is obtained in cases (a) and (b) in Theorem 4.6 by setting i=αβ and γ, respectively and the definitive choice we make for an ωj-covering group for G is H(αβ+j) and H(γ+j), respectively, where [ωj] has order pj and 0jβγ.

Theorem 4.7.

Let G=Gp(α,β,ϵ,γ) with α<β+ϵ. Define the sequence

  1. H(i)=Gp(αϵ+i,β,i,γϵ+i) for i=0,,αγ if γ>ϵ;

  2. H(i)=Gp(αγ+i,β,ϵγ+i,i) for i=0,,αϵ if γϵ.

Then H(i) is a metacyclic p-group and |M(H(i))|=pαγi in case (a) and pαϵi in case (b). Suppose [ω]M(H(i)) of order pj. Then H(i+j) is an ω-covering group for H(i), in particular the last group in each sequence is a Schur group that is a representation group for all the groups in the sequence.

Proof.

In case (a), αϵ+iαϵ>0. By definition γϵ+i>0, also β(γϵ+i)β(αϵ)>0, (αϵ+i)(γϵ+i)=αγ and β+i(αϵ+i)>0. Thus H(i) is a metacyclic p-group from Lemma 3.2, |M(H(i)|=pαγi from Corollary 4.2 and in particular H(αγ) is a Schur group.

Now let H(i+j)=x,y. Then by Lemma 4.5, Z(H(i+j))=xpα(ϵ+i+j),ypα(ϵ+i+j) and H(i+j)=xpαγ. Therefore, Z(H(i+j))H(i+j)=xpk, where k=max{α(ϵ+i+j),αγ}max{αϵ,αγ}=αϵ. Thus, Ai=xpαϵ+iZ(H(i+j))H(i+j). Next set a=xxpαϵ+i and b=yxpαϵ+i and observe that H(i+j)/AiH(i). Finally, |Ai|=p(αϵ+i+j)(αϵ+i)=pj=o([ω]) and the proof is completed as in that of Theorem 4.6.

In case (b) the argument is now essentially the same as in case (a), but with ϵ and γ interchanged. □

The group G=Gp(α,β,ϵ,γ) is obtained in cases (a) and (b) in Theorem 4.7 by setting i=αγ and αϵ respectively and the definitive choice we make for an ωj-covering group for G is H(αγ+j) and H(αϵ+j) respectively, where [ωj] has order pj and 0jα(ϵ+γ).

5 Regular elements and projective characters of metacyclic p-groups of positive type

We first aim to find the ωi-regular elements of a metacyclic p-group G of positive type, where ωi is a 2-cocycle of G with o([ωi])=pi in M(G). The first result is a slight generalization of [10, Lemma 4]. The original result is stated for the case when ϵ = 0, but the proof of Theorem 4 of the same paper shows that this result holds for any metacyclic p-group of positive type.

Lemma 5.1.

Let H=Gp(α,β,ϵ,γ)=x,y. Set z=xpα1 and let G=H/z. Then hz,hzGZ(G) are conjugate in G if and only if h and h are conjugate in H.

For notational convenience we now reset the indices in Theorems 4.6 and 4.7, so that H(0)=Gp(α,β,ϵ,γ).

Corollary 5.2.

Let GGp(α,β,ϵ,γ) with M(G) non-trivial and let [ω]M(G) with [ω][1]. Then all the ω-regular elements of G are central.

Proof.

Using (the now revised indices) we set H(1)=x,y=Gp(α+1,β,ϵ,γ+1) in Theorem 4.6 and Gp(α+1,β,ϵ+1,γ+1) in Theorem 4.7. Then from the two theorems A1=z=xpα and H(1) is an ω1-covering group for G, where [ω1]M(G) with o([ω1])=p.

Now Lemma 5.1 yields that hz,hzGZ(G) are conjugate in G if and only if h and h are conjugate in H(1). In particular, h is conjugate to hzk in H(1) for all kN and for all hzGZ(G). Next, if we take any χIrr(H(1)|λ) with tra(λ)=[ω1], then λ(z)=e2πik/p for some k{1,,p1}. However this means that χ(h)=χ(hz)=χ(h)λ(z). In particular, since λ(z)1, we must have χ(h)=0, and thus hz is not ω1-regular. Then since M(G) is cyclic every non-trivial [δ]M(G) satisfies [δ]n=[ω1] for some nN. Therefore, GδGδn=Gω1 by Lemma 2.3 and so the result for [ω1] extends to all nontrivial elements of M(G). □

Corollary 5.3.

Let GGp(α,β,ϵ,γ) and let [ωi]M(G) of order pi for i=0,,vp(|M(G)|). Then Zωi(G)apγ+i,bpγ+i.

Proof.

Let H(i)=x,y be the ωi-covering group of G=H(0) (with indices reset) as given in Theorems 4.6 and 4.7. Then by [12, Theorem 3.5], Z(H(i))/xpαZωi(G). Furthermore Z(H(i))=xpγ+i,ypγ+i from Lemma 4.5. So writing Ai=xpα we obtain Zωi(G)xpγ+i,ypγ+i/Ai=xpγ+iAi,ypγ+iAiapγ+i,bpγ+i.

Our second aim in this section is to describe the irreducible ωi-characters of G=Gp(α,β,ϵ,γ) for the 2-cocycles ωi described below. This will be achieved by describing the elements of Proj(Zi,(ωi)Zi), where Zi=Zωi(G), and by finding the degrees of the elements of Proj(G,ωi).

To achieve the first aim we need to choose a transversal R={r(g):gG} for Ai in H(i), where Ai and H(i)=x,y are as described in Theorems 4.6 and 4.7 with indices reset so that H(0)=G. A natural choice for the elements of R in Z(H(i)) is to define r(apγ+isbpγ+it)=xpγ+isypγ+it for all s{0,,pα(γ+i)1} and for all t{0,,pβ(γ+i)1}. Now let λi be the element of Irr(Ai) with trai(λi)=ωi so that both ωi and [ωi] have order pi as explained in Section 1. Define ζi=λi(xpα), so that ζi is a primitive pith root of unity.

Lemma 5.4.

With notation as above Proj(Zi,(ωi)Zi)={δμ:μIrr(Zi)} with δ(apγ+isbpγ+it)=ζs,for s{0,,pα(γ+i)1},t{0,,pβ(γ+i)1}, and where ζ is a pαγth root of unity with ζpα(γ+i)=ζi.

Proof.

Let Ti be the subgroup of Z(H(i)) which is mapped to Zi under the epimorphism πi:H(i)G with kernel Ai. Then Ti=xpγ+i,ypγ+iCpα(γ+i)×Cpβ(γ+i). So we just need to find one ηIrr(Ti) satisfying η(xpγ+i)pα(γ+i)=η(xpα)=ζi. One such character η is defined by η(xpγ+i)=ζ, where ζpα(γ+i)=ζi and η(ypγ+i)=1. Finally δ can be constructed by defining δ(apγ+isbpγ+it)=η(r(apγ+isbpγ+it))=η(xpγ+isypγ+it)=ζs.

Finally Proj(Zi,(ωi)Zi)={δμ:μIrr(Zi)}. However, Zi=apγ+i,bpγ+iCppα(γ+i)1×Cppβ(γ+i)1.

We now observe from Corollary 5.2 and Proposition 2.2, that ξ(1)2=|G:Zi|=p2(γ+i1) for all ξProj(G,ωi) for i0. Secondly we have ξ(g)=0 for all gZi and all ξProj(G,ωi) for i0. Thirdly there is a bijection between the elements of Proj(G,ωi) and the elements of Irr(Zi) defined by ξμ if ξ(g)=ξ(1)δ(g)μ(g) for all gZi, where δ is as described in Lemma 5.4 and i0. We thus have a complete description of the elements of Proj(G,ωi) for i0; however, the exact values of δ depend upon the initial choice of representation group that we made and the consequent choice of transversal R. The differences these choices make are discussed in detail in [Citation4], but in essence it means that δ is only defined up to certain roots of unity.

6 Metacyclic 2-groups of negative type

Let G be a metacyclic 2-group of negative type. In this section we will find M(G), a representation group for G that is a Schur group, |Proj(G,ω)| and the degree of each element of Proj(G,ω) for [ω] non-trivial in M(G).

We first compute the Schur multipliers of all groups of negative type.

Proposition 6.1.

Let GG2(α,β,ϵ,γ,). Then

  1. if ϵ = 0 and β>γ, then M(G)C2;

  2. if ϵ=1, or ϵ = 0 and β=γ, then M(G) is trivial.

Proof.

We use Theorem 4.1 and its notation. Converting notations, we have m=2α,s=2β,r=2αγ1, and t=2αϵ. We see that (r1,m)=(2αγ2,2α)=2,v2(1+r++rs1)=αγ+β1 from [1, Lemma 2.3] and αγ+β1α1 with equality if and only if β=γ. Now overall we have n=2l/2α=l/2α1, where l=(2αγ+β1,2αϵ). Now if ϵ = 0 and β>γ, then l=2α. On the other hand if ϵ=1, or ϵ = 0 and β=γ, then l=2α1.

We next aim to find a representation group, that is also a Schur group, for GG2(α,β,ϵ,γ,) when G has nontrivial Schur multiplier, that is when ϵ = 0 and β>γ. One method of constructing such a representation group, that is also a Schur group, can be seen in the proof of [3, Theorem IV.2.10]. More specifically, we start with any metacyclic group G=a,b:am=1,bs=at,bab1=ar,where k=λm/(m,r1) (with λ as defined in Section 3). We then find an integer u satisfying (u,m)=1 and u(r1)/(m,r1)(modm/(m,r1)) and set vu(m,r1)+1(modmλ) with vN and vmλ+1. Then a representation group for G is given by H=x,y:xmλ=1,ys=xw,yxy1=xv, where w=mλ/(mλ,r1).

Theorem 6.2.

Let GG2(α,β,0,γ,) with β>γ. Then H=G2(α+1,β,1,γ+1,) is a representation group for G and a Schur group.

Proof.

Using the notation above m=2α,s=2β,r=2αγ1,λ=2,m/(m,r1)=2α1 and (r1)/(m,r1)=2αγ11. Set u=2αγ11,v=2αγ1 and w=2α to obtain the representation group and Schur group H=x,y:x2α+1=1,y2β=x2α,yxy1=x2αγ1=G2(α+1,β,1,γ+1,).

The remainder of this section will be concerned with finding the degrees of the irreducible ω-characters of G, where [ω][1]. We will also see why the approach of Section 4 does not work for metacyclic 2-groups that are only of negative type. Both of these aims will be achieved through finding the number of irreducible ω-characters of G.

Now we have shown that every metacyclic 2-group G of negative type with non-trivial Schur multiplier has M(G)C2. So for any representation group H of G, Irr(H)=Irr(H|λ1)̇Irr(H|λ2), where λ1 and λ2 are the trivial and nontrivial character of C2 respectively. Thus, |Irr(H|λ1)|+|Irr(H|λ2)|=k(H), the number of conjugacy classes of H. However, |Irr(H|λ1)|=|Irr(G)|=k(G), and therefore if ω is a 2-cocycle of G with [ω][1], then |Proj(G,ω)|=|Irr(H|λ2)|=k(H)k(G). These numbers can mainly be obtained from [11, Theorems 3.1 and 3.4]. For convenience, we now summarize these results.

Lemma 6.3.

Let GG2(α,β,ϵ,γ,). Then k(G)=(2γ1)2α+β2γ1+2α+βγ2+3(2β1).

Proof.

For G of nilpotency class at least three this follows from the aforementioned result and [1, Proposition 3.7], since the same number is obtained for k(G) if γ = 0 or 1. We just have to check that the results hold when G has nilpotency class one or two. Now if G is abelian, we have that α = 1 and γ = 0 and so the formula for k(G) gives 2β+1=|G|. Now for G of nilpotency class two we have that α = 2 and γ = 0, so that |G/Z(G)|=22 from [1, Proposition 2.5] and hence k(G)=5(2β1)=2β+3(2β1) from [10, Lemma 5]. □

Theorem 6.4.

Let GG2(α,β,0,γ,) with β>γ. Let ω be a 2-cocycle of G with [ω][1]. Then |Proj(G,ω)|=2α+β2(γ+1).

Proof.

We need to find k(H)k(G), where H is as described in Theorem 6.2. Now k(H)k(G)=(2γ+11)2α+β2γ2(2γ1)2α+β2γ1=(2γ+11)2α+β2γ2(2γ+12)2α+β2γ2=2α+β2(γ+1).

Proposition 6.5.

Let GG2(α,β,0,γ,) with β>γ. Let ω be a 2-cocycle of G with [ω][1]. Then Zω(G)b2γ+1.

Proof.

Let H=x,y be the representation group for G defined in Theorem 6.2. Then by [1, Proposition 2.5] we have Z(H)=Z(G2(α+1,β,1,γ+1,))=x2α,y2γ+1. So from [12, Theorem 3.5] Zω(G)Z(H)/x2αb2γ+1.

Note that if G is a non-abelian group in Proposition 6.5 then there are non-central ω-regular elements of G. To see this, by Theorem 6.4 the total number of ω-regular conjugacy classes of G is 2α+β2(γ+1). However Proposition 6.5 yields that 2β/2γ+1=2β(γ+1) of these ω-regular conjugacy classes are central. This leaves a total of 2α+β2(γ+1)2β(γ+1)=2β(γ+1)(2α(γ+1)1) ω-regular conjugacy classes that are non-central. This number is nonzero since α>γ+1 for all non-abelian metacyclic 2-groups of negative type. This observation implies that a metacyclic 2-group of nilpotency class two that is of both positive and negative type must have trivial Schur multiplier. Thus, this also offers an alternative proof that the groups G2(2,β,0,0,) for βN are of negative type only.

Theorem 6.6.

Let GG2(α,β,0,γ,) with β>γ. Let ω be a 2-cocycle of G with [ω][1]. Then ξ(1)=2γ+1 for all ξProj(G,ω).

Proof.

H=G2(α+1,β,1,γ+1,) is a representation group for G by Theorem 6.2. Let N=x2,y2γ, so that N has index 2γ+1 in H. Now y2γx2y2γ=x2(2αγ1)2γand v2((2αγ1)2γ1)=α from [1, Lemma 2.3], so that 2(2αγ1)2γ2(mod2α+1) and hence N is an abelian subgroup of H.

From the presentation of H, it is clear that yx2y1N. We also have x1y2γx=x(2αγ1)2γ1y2γN,and so N is a normal subgroup of H. Now χ(1) divides |H:N| for all χIrr(H) by Ito’s theorem (see [5, Theorem 6.5.6]). So in particular ξ(1) divides 2γ+1 for all ξProj(G,ω). Now using Theorem 6.4 |G|=2α+β=(2γ+1)2|Proj(G,ω)|=ξProj(G,ω)ξ(1)2.

Thus, the only possibility is ξ(1)=2γ+1 for all ξProj(G,ω). □

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