Schur multipliers of special π-groups of rank 2

• Sumana Hatui
From the journal Journal of Group Theory

Abstract

Let G be a special p-group with center of order p2. Berkovich and Janko asked to find the Schur multiplier of G in [Y. Berkovich and Z. Janko, Groups of Prime Power Order. Volume 3, De Gruyter Exp. Math. 56, Walter de Gruyter, Berlin, 2011; Problem 2027]. In this article, we answer this question by explicitly computing the Schur multiplier of these groups.

1 Introduction

Let G be a finite p-group, and let dβ’(G) denote the cardinality of a minimal generating set of G. The commutator subgroup and center of G are denoted by Gβ² and Zβ‘(G) respectively. By ESpkβ’(pn), we denote the extraspecial p-group of order pn of exponent pk, and by β€pk, we denote the elementary abelian p-group of rank k for kβ₯1. For p=2, there are two extraspecial p-groups of order 8 up to isomorphism, both of exponent 4; one is the quaternion group Q8 which has trivial Schur multiplier and the other is the dihedral group D8 which has Schur multiplier of order 2.

A finite p-group G is called a special p-group of rank k if Gβ²=Zβ‘(G) is elementary abelian of order pk and G/Gβ² is elementary abelian. A group G is called a capable group if there exists a group H such that GβH/Zβ‘(H). We denote the epicenter of a group G by Z*β’(G), which is the smallest central subgroup K of G such that G/K is capable.

The Schur multiplier of a group G, denoted by Mβ‘(G), is the second homology group H2β‘(G,β€), which was introduced by Schur in his work studying projective representations of groups. The special p-groups of minimum rank are the extraspecial p-groups, and their Schur multiplier was studied in [3]. The Schur multiplier of special p-groups having maximum rank was studied in [12]. In this article, we determine the Schur multiplier of special p-groups of rank 2, giving an answer to the question that was asked by Berkovich and Janko in [1, Problem 2027].

We state our main results now. The following result describes the Schur multiplier of G when Gp=Gβ².

Theorem 1.1.

Let G be a special p-group of rank 2 with d=dβ’(G) and Gp=Gβ². Then the following assertions hold:

1. Either Z*β’(G)=Zβ‘(G) or G is capable.

2. Mβ‘(G) is elementary abelian of order p12β’dβ’(d-1)-2 iff Z*β’(G)=Zβ‘(G).

3. Mβ‘(G) is of order p12β’dβ’(d-1)-1 and exponent at most p2 iff G is capable.

4. For every central subgroup Z of order p, G/Z is isomorphic to

ESp2(p2β’m+1)Γβ€pd-2β’mβ(mβ₯2),ESp2β’(p3)Γβ€pd-2βππβQ8Γβ€2d-2.

By [5, Proposition 3] and Theorem 1.1, we have the immediate corollary.

Corollary 1.2.

If G is a special p-group of rank 2 of order pn (nβ₯8) with Gp=Gβ², then Mβ‘(G) is elementary abelian of order p12β’(n-2)β’(n-3)-2.

The following result describes the Schur multiplier of G when Gp is cyclic of prime order.

Theorem 1.3.

If G is a special p-group of rank 2 with d=dβ’(G) and GpβZp, then the following assertions hold:

1. G is not capable, and either Z*β’(G)=Zβ‘(G) or Z*β’(G)=Gp.

2. Mβ‘(G) is elementary abelian.

3. The following are equivalent:

1. Mβ‘(G) is of order p12β’dβ’(d-1)-2.

2. Z*β’(G)=Zβ‘(G).

3. G/GpβESpβ’(p2β’m+1)Γβ€pd-2β’m, mβ₯2.

4. The following are equivalent:

1. Mβ‘(G) is of order p12β’dβ’(d-1).

2. Z*β’(G)=Gp.

3. G/GpβESpβ’(p3)Γβ€pd-2.

Now we are left only one case when Gp is trivial.

Theorem 1.4.

If G is a special p-group of rank 2 with d=dβ’(G) and Gp=1, p odd, then the following assertions hold:

1. Mβ‘(G) is elementary abelian.

2. p12β’dβ’(d-1)-2β€|Mβ‘(G)|β€p12β’dβ’(d-1)+3.

3. G is capable if and only if G is isomorphic to one of the following groups:

Ξ¦4β’(15)=γΞ±,Ξ±1,Ξ±2,Ξ²1,Ξ²2β£β’[Ξ±i,Ξ±]=Ξ²i,Ξ±p=Ξ±ip=Ξ²ip=1(i=1,2)γ,
Ξ¦12β’(16)=ESpβ’(p3)ΓESpβ’(p3),
Ξ¦13β’(16)=γΞ±1,Ξ±2,Ξ±3,Ξ±4,Ξ²1,Ξ²2β£β’[Ξ±i,Ξ±i+1]=Ξ²i,[Ξ±2,Ξ±4]=Ξ²2,Ξ±ip=Ξ±3p=Ξ±4p=Ξ²ip=1(i=1,2)γ,
Ξ¦15β’(16)=γΞ±1,Ξ±2,Ξ±3,Ξ±4,Ξ²1,Ξ²2β£β’[Ξ±i,Ξ±i+1]=Ξ²i,[Ξ±3,Ξ±4]=Ξ²1,
ββ[Ξ±2,Ξ±4]=Ξ²2g,Ξ±ip=Ξ±3p=Ξ±4p=Ξ²ip=1(i=1,2)γ,
T=γx1,β¦,x5,c1,c2β£β’[x2,x1]=[x5,x3]=c1,[x3,x1]=[x5,x4]=c2,xip=cjp=1,1β€iβ€5,β1β€jβ€2γ.

4. |Mβ‘(G)|=p12β’dβ’(d-1)+3 if and only if G is isomorphic to Ξ¦4β’(15).

5. |Mβ‘(G)|=p12β’dβ’(d-1)+2 if and only if G is isomorphic to Ξ¦12β’(16), Ξ¦13β’(16) or Ξ¦15β’(16).

6. |Mβ‘(G)|=p12β’dβ’(d-1)-1 if and only if G is isomorphic to T.

7. Mβ‘(G) is of order p12β’dβ’(d-1)-2 if and only if Z*β’(G)=Zβ‘(G). In this case, G/ZβESpβ’(p2β’m+1)Γβ€pd-2β’m, mβ₯2, for every central subgroup Z of order p.

8. Mβ‘(G) is of order p12β’dβ’(d-1) if and only if Z*β’(G)ββ€p. In this case,

G/Z*β’(G)βESpβ’(p3)Γβ€pd-2.

The following result is for p=2.

Theorem 1.5.

Let G be a special 2-group of rank 2. Then G2=Gβ² holds.

2 Preliminaries

For a finite group G of class 2 with G/Gβ² elementary abelian, the following construction is given in [3]. We consider G/Gβ² and Gβ² as vector spaces over π½p, which we denote by V,W respectively. The bilinear map (-,-):VΓVβW is defined by (v1,v2)=[g1,g2] for v1,v2βV such that vi=giβ’Gβ²,iβ{1,2} for some g1,g2βG. Let X1 be the subspace of VβW spanned by all

v1β(v2,v3)+v2β(v3,v1)+v3β(v1,v2)

for v1,v2,v3βV. Consider a map f:VβW given by fβ’(gβ’Gβ²)=gp for gβG. Let X2 be the subspace spanned by all vβfβ’(v), vβV, and take X:=X1+X2. Now consider a homomorphism Ο:Vβ§Vβ(VβW)/X given by

Ο(v1β§v2)=(v1βf(v2)+(2p)v2β(v1,v2))+X.

There exists an abelian group M* with a subgroup N isomorphic to (VβW)/X, such that

1βNβM*β’βπβ’Vβ§Vβ1

is exact and

Οβ’ΞΎβ’(Ξ±)=Ξ±pβforβ’Ξ±βM*.

Now we consider a homomorphism Ο:Vβ§VβW given by

Οβ’(v1β§v2)=(v1,v2)

for all v1,v2βV. Notice that Ο is an epimorphism. We let M be the subgroup of M* containing N such that M/NβKerβ‘Ο. We use this notation throughout the paper without further reference.

With the above setting, we have the following theorem.

Mβ‘(G)βM.

Note.

It is easy to observe that X1 is generated by the set

{xΒ―1β[x2,x3]+xΒ―2β[x3,x1]+xΒ―3β[x1.x2]β£x1,x2,x3βS},

where S is a set of generators of G and xΒ― is the image of x in G/Gβ².

Suppose G has a free presentation F/R. Let Z=S/R be a central subgroup of G. Then the map from (F/Fβ²β’R)Γ(S/R) to (Fβ²β©R)/[F,R] defined by

(xβ’Fβ²β’R,sβ’R)β¦[x,s]β’[F,R]

is a well-defined bilinear map and induces a homomorphism

Ξ»Z:(G/Gβ²)βZβMβ‘(G),

called the Ganea map.

Theorem 2.2 ([4]).

Let Z be a central subgroup of a finite group G. Then the following sequence is exact:

Theorem 2.3 ([2]).

Let Z be a central subgroup of a finite group G. Then we have ZβZ*β’(G) if and only if (G/Gβ²)βZ=Kerβ‘Ξ»Z.

By [8, Corollary 3.2.4], we have X=Kerβ‘Ξ»Zβ‘(G). Hence, by Theorem 2.3, we have the following result.

Lemma 2.4.

Let Z be a central subgroup of a group G of nilpotency class 2. Then ZβZ*β’(G) if and only if (G/Gβ²)βZ is contained in X.

Let v1,v2,β¦,vd be the generators of V such that {fβ’(v1),fβ’(v2),β¦,fβ’(vr)} is a basis of Gp. Then the set

{viβf(vi),vlβf(vi),(viβf(vj)+vjβf(vi))β£1β€i<jβ€r,(r+1)β€lβ€d}

forms a basis of X2, from which the following result follows.

Proposition 2.5 ([12, Proposition 3.3]).

Let G be a special p-group, d=dβ’(G), and Gp of order r. Then |X2|=prβ’d-12β’rβ’(r-1).

By Theorem 2.2, we have

As X=Kerβ‘Ξ»Zβ‘(G), so |Imβ‘Ξ»Zβ‘(G)|=p2β’d|X|. Hence, by (2.1), taking Z=Zβ‘(G), we have

(2.2)|Mβ‘(G)|=p12β’dβ’(d-1)-2.p2β’d|X|

Now we recall the following results which will be used in the proof of the main results.

Theorem 2.6 ([9, Main Theorem]).

Let G be a p-group of order pn. Then we have |Mβ‘(G)|=p12β’(n-1)β’(n-2)+1 if and only if GβESpβ’(p3)ΓZpn-3.

Theorem 2.7 ([10, Theorem 21]).

Let G be a p-group of order pn. Then we have |Mβ‘(G)|=p12β’(n-1)β’(n-2) if and only if G is isomorphic to one of the following groups:

1. Gββ€p2Γβ€pn-2,

2. GβD8Γβ€2n-3,

3. Gββ€p4ββ€p (pβ 2).

Theorem 2.8 ([11, Theorem 11]).

Let G be a group of order pn such that G/Gβ² is elementary abelian of order pn-1. Then |Mβ‘(G)|=p12β’(n-1)β’(n-2)-1 if and only if G is isomorphic to one of the following groups:

1. ESp2β’(p3)Γβ€pn-3,

2. Q8Γβ€2n-3,

3. ESβ’(p2β’m+1)Γβ€pn-2β’m-1 (mβ₯2).

3 Proofs

In this section, we prove our main results.

Proof of Theorem 1.1.

Consider Gp=Gβ²ββ€pΓβ€p. Now, by Proposition 2.5, |X2|=p2β’d-1. Let v1,v2βV such that {fβ’(v1),fβ’(v2)} is a basis of W. Observe that X2 is generated by the set

{viβfβ’(vi),vjβfβ’(vi),(v1βfβ’(v2)+v2βfβ’(v1))β£i=1,2β’andβ’β3β€jβ€d}.

Hence p2β’dβ₯|X|β₯p2β’d-1, and by (2.2),

p12β’dβ’(d-1)-2β€|Mβ‘(G)|β€p12β’dβ’(d-1)-1.

(a) Observe that |X|=p2β’d, i.e., the set {v1βfβ’(v2),v2βfβ’(v1)} is contained in X if and only if Z*β’(G)=Zβ‘(G), follows by Lemma 2.4. Another possibility is |X|=p2β’d-1, i.e., X1βX2 if and only if v1βfβ’(v2),v2βfβ’(v1) are not in X. Hence, by Lemma 2.4, Z*β’(G)=1. So G is capable.

(b) By (2.2), |X|=p2β’d if and only if

|Mβ‘(G)|=p12β’dβ’(d-1)-2.

By Theorem 2.3 and Theorem 2.2, taking Z=Zβ‘(G), we see that Mβ‘(G) embeds in Mβ‘(G/Zβ‘(G)) which is elementary abelian. So, in this case, Mβ‘(G) is elementary abelian.

(c) For xβMβ‘(G), xpβ(VβW)/X and VβW/X is elementary abelian, so xp2=1.

By (2.2), X1βX2 i.e., |X|=p2β’d-1 if and only if, by (2.2),

|Mβ‘(G)|=p12β’dβ’(d-1)-1.

Hence, by Lemma 2.4, Z*β’(G)=1, so G is capable. The converse follows from (b).

(d) Observe that, in both the cases, |Mβ‘(G/Z)|=p12β’dβ’(d-1)-1 follows from (2.1), taking Z a central subgroup of order p. Therefore, by Theorem 2.8,

G/ZβESp2(p2β’m+1)Γβ€pd-2β’mβ(mβ₯2),ESp2β’(p3)Γβ€pd-2βorβQ8Γβ€2d-2.β

Proof of Theorem 1.3.

(a) Assume Gp is cyclic of order p. Now, by Proposition 2.5, |X2|=pd. Let v1βV such that Gp=γfβ’(v1)γ. Observe that X2 is generated by the set

Mβ²:={viβf(v1),β1β€iβ€d}.

Hence, by Lemma 2.4, GpβZ*β’(G). Hence G is not capable with Z*β’(G)=Gp or Z*β’(G)=Zβ‘(G).

Using Theorem 2.3 and taking Z=Gp in (2.1), we have

(3.1)|Mβ‘(G)|=|Mβ‘(G/Gp)|p

Now let v1,v2βV such that (v1,v2)βGβ²βGp. Then the set

Nβ²:={viβ(v1,v2)+v1β(v2,vi)+v2β(vi,v1)β£3β€iβ€d}

is linearly independent in X1, so |X1|β₯pd-2. The set Mβ²βͺNβ² is linearly independent in X and Mβ²β©Nβ²=β. Thus

p2β’dβ₯|X|β₯p2β’d-2.

Hence, by (2.2),

p12β’dβ’(d-1)-2β€|Mβ‘(G)|β€p12β’dβ’(d-1).

(c) Now similarly, as described in the proof of Theorem 1.1, Z*β’(G)=Zβ‘(G) if and only if |Mβ‘(G)|=p12β’dβ’(d-1)-2, i.e., by (3.1),

|Mβ‘(G/Gp)|=p12β’dβ’(d-1)-1,

which happens if and only if

G/GpβESpβ’(p2β’m+1)Γβ€pd-2β’m,mβ₯2,

follows from Theorem 2.8.

(d) By Theorem 2.7, it follows that there is no G/Gp such that

|Mβ‘(G/Gp)|=p12β’dβ’(d-1).

Thus, by (3.1), |Mβ‘(G)| cannot be of order p12β’dβ’(d-1)-1. Hence

|Mβ‘(G)|=p12β’dβ’(d-1)

if and only if Z*β’(G)=Gp. By (3.1),

|Mβ‘(G/Gp)|=p12β’dβ’(d-1)+1,

which happens if and only if

G/GpβESpβ’(p3)Γβ€pd-2,

follows by Theorem 2.6.

(b) By (c) and (d), it follows that p must be odd. The group G/Gp is of exponent p and p odd, so the homomorphism Ο, described in Section 2, is the trivial map, and therefore Οβ’ΞΎβ’(x)=xp=1 for xβMβ‘(G/Gp). Thus Mβ‘(G/Gp) is elementary abelian. Since GpβZ*β’(G), by Theorem 2.3 and Theorem 2.2, Mβ‘(G) embeds in Mβ‘(G/Gp). Therefore, Mβ‘(G) is also elementary abelian. The proof is complete now. β

Proof of Theorem 1.4.

(a) Since p is odd and Gp=1, the homomorphism Ο, described in Section 2, is the trivial map, and therefore Οβ’ΞΎβ’(x)=xp=1. Thus Mβ‘(G) is elementary abelian.

Let z,zβ² be the generators of Zβ‘(G), and let x1,x2,β¦,xd be the generators of G such that [x1,x2]βγzγ is non-trivial. Then the set

A:={xiβ[x1,x2]+x1β[x2,xi]+x2β[xi,x1]β£3β€iβ€d}

consists of d-2 linearly independent elements of X1.

Now if, for some xkβ{3,4,β¦,d}, [x1,xk]βγzβ²γ is non-trivial, then the set

B:={xiβ[x1,xk]+x1β[xk,xi]+xkβ[xi,x1]β£3β€iβ€d,iβ k}

consists of d-3 linearly independent elements of X1. Thus AβͺB consists of (2β’d-5) linearly independent elements of X1.

If we have 1β [x2,xk]βγzβ²γ for some kβ{3,4,β¦,d}, then a similar conclusion holds. Suppose then that [x1,xk] and [x2,xk] are all trivial or in γzγ for all kβ{3,4,β¦,d}. Say, [x3,x4]βγzβ²γ. In this case,

B1:={xiβ[x3,x4]+x3β[x4,xi]+x4β[xi,x3]β£1β€iβ€d,iβ 3,4}

consists of d-2 independent elements of X1. Thus AβͺB1 consists of 2β’d-4 linearly independent elements of X1.

Hence, in both cases, 2β’dβ₯|X|=|X1|β₯p2β’d-5 holds. By (2.2), it follows that

p12β’dβ’(d-1)-2β€|Mβ‘(G)|β€p12β’dβ’(d-1)+3,

which proves (b).

Now if G is not capable, then, by (2.1) and [9, Main Theorem], we have

p12β’dβ’(d-1)-2β€|Mβ‘(G)|=|Mβ‘(G/Z)|pβ€p12β’dβ’(d-1)βforβ’ZβZ*β’(G),Zββ€p.

By Theorem 2.7, there is no G and central subgroup Z such that

|Mβ‘(G/Z)|=p12β’dβ’(d-1).

Hence

(3.2)|Mβ‘(G)|=p12β’dβ’(d-1)-2βorβp12β’dβ’(d-1)βifβ’Gβ’is not capable.

Assume then that G is capable. By [5, Proposition 3], p5β€|G|β€p7, with p odd. If |G|=p5, then, looking through the list of groups given in [7], it follows that GβΞ¦4β’(15). Since |X|=p, it follows by (2.2) that

|Mβ‘(G)|=p6=p12β’dβ’(d-1)+3.

If |G|=p6, then, looking through the list of groups given in [7], it follows that GβΞ¦12β’(16), Ξ¦13β’(16) or Ξ¦15β’(16). Since |X|=p4, it follows by (2.2) that |Mβ‘(G)|=p8=p12β’dβ’(d-1)+2.

Now consider groups of order p7 of exponent p. By [6], it follows that there is only one capable group

G=γx1,β¦,x5,c1,c2β£[x2,x1]=[x5,x3]=c1,[x3,x1]=[x5,x4]=c2,xip=cjp=1,β1β€iβ€5,β1β€jβ€2γ

up to isomorphism. By (2.2), |Mβ‘(G)|=p9=p12β’dβ’(d-1)-1.

Now (c), (d), (e), (f) follow by (3.2).

(g) By (2.1), we have Z*β’(G)βZβ‘(G) if and only if |Mβ‘(G)|=p12β’dβ’(d-1)-2, taking Z=Zβ‘(G). Hence, by (2.1), we have that, for every central subgroup Z of order p, |Mβ‘(G/Z)|=p12β’dβ’(d-1)-1, and thus

G/ZβESpβ’(p2β’m+1)Γβ€pd-2β’m,mβ₯2,

follows from Theorem 2.8.

(h) Suppose Z*β’(G)ββ€p. By (3.2) and (g), it follows that |Mβ‘(G)|=p12β’dβ’(d-1).

Conversely, suppose |Mβ‘(G)|=p12β’dβ’(d-1). From the previous cases, it follows that G is not capable since Z*β’(G)ββ€p.

In this case, we have

|Mβ‘(G)|=|Mβ‘(G/Z*β’(G))|p.

By Theorem 2.6, it follows that |Mβ‘(G/Z*β’(G))|=p12β’dβ’(d-1)+1 if and only if

G/Z*β’(G)βESpβ’(p3)Γβ€pd-2.

The proof is complete now. β

Proof of Theorem 1.5.

By Theorem 1.3, it follows that there is no special 2-group of rank 2 with G2ββ€2.

Assume that p=2. As in the proof of Theorem 1.4, we conclude that

p12β’dβ’(d-1)-2β€|Mβ‘(G)|β€p12β’dβ’(d-1)+3.

Let Z be a central subgroup of order 2. If G is not capable, then, by (2.1),

212β’dβ’(d-1)-2β€|Mβ‘(G)|=|Mβ‘(G/Z)|2β€212β’dβ’(d-1)βforβ’ZβZ*β’(G).

By Theorems 2.6, 2.7 and 2.8, there is no group G and central subgroup Z such that G/Z of exponent 2 and 212β’dβ’(d-1)-1β€|Mβ‘(G/Z)|β€212β’dβ’(d-1)+1. Hence G must be capable, and |Mβ‘(G/Z)|β€212β’dβ’(d-1)-2.

Suppose |Mβ‘(G)|=212β’dβ’(d-1)-2+k, 0β€kβ€5. Then, by (2.2),

|Kerβ‘Ξ»Zβ‘(G)|=|X|=22β’d-k.

Hence 1β€|Imβ‘Ξ»Z|=2mβ€2k. By (2.1), it follows that

212β’dβ’(d-1)-2+k-m=|Mβ‘(G)||Imβ‘Ξ»Z|=|Mβ‘(G/Z)|2β€212β’dβ’(d-1)-3.

Hence k-mβ€-1, which is not possible. So there is no special 2-group of rank 2 with G2=1. The proof is complete now. β

Communicated by Christopher W. Parker

Funding statement: The research of the author is partly supported by Infosys grant.

References

[1] Y. Berkovich and Z. Janko, Groups of Prime Power Order. Volume 3, De Gruyter Exp. Math. 56, Walter de Gruyter, Berlin, 2011. Search in Google Scholar

[2] F.βR. Beyl, U. Felgner and P. Schmid, On groups occurring as center factor groups, J. Algebra 61 (1979), no. 1, 161β177. 10.1016/0021-8693(79)90311-9Search in Google Scholar

[3] N. Blackburn and L. Evens, Schur multipliers of p-groups, J. Reine Angew. Math. 309 (1979), 100β113. 10.1515/crll.1979.309.100Search in Google Scholar

[4] T. Ganea, Homologie et extensions centrales de groupes, C. R. Acad. Sci. Paris SΓ©r. A-B 266 (1968), A556βA558. Search in Google Scholar

[5] H. Heineken, Nilpotent groups of class two that can appear as central quotient groups, Rend. Semin. Mat. Univ. Padova 84 (1990), 241β248. Search in Google Scholar

[6] H. Heineken, L.βC. Kappe and R.βF. Morse, On the classification of special p-groups of rank two that appear as central quotient groups, preprint. Search in Google Scholar

[7] R. James, The groups of order p6 (p an odd prime), Math. Comp. 34 (1980), no. 150, 613β637. 10.1090/S0025-5718-1980-0559207-0Search in Google Scholar

[8] G. Karpilovsky, The Schur Multiplier, London Math. Soc. Monogr. (N.βS.) 2, The Clarendon Press, Oxford University Press, New York, 1987. Search in Google Scholar

[9] P. Niroomand, On the order of Schur multiplier of non-abelian p-groups, J. Algebra 322 (2009), no. 12, 4479β4482. 10.1016/j.jalgebra.2009.09.030Search in Google Scholar

[10] P. Niroomand, A note on the Schur multiplier of groups of prime power order, Ric. Mat. 61 (2012), no. 2, 341β346. 10.1007/s11587-012-0134-4Search in Google Scholar

[11] P. Niroomand, Classifying p-groups by their Schur multipliers, Math. Rep. (Bucur.) 20(70) (2018), no. 3, 279β284. Search in Google Scholar

[12] P.βK. Rai, On the Schur multiplier of special p-groups, J. Pure Appl. Algebra 222 (2018), no. 2, 316β322. 10.1016/j.jpaa.2017.04.004Search in Google Scholar