# Finitep-Group with Small Abelian Subgroups

Authors: Yuemei Mao; Qianlu Li
Journal: Journal of Mathematics
 (2013)

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## Abstract

A finitep-group G is said to have the property P, if, for any abelian subgroup M of G, there is |MZ(G)/Z(G)|≤p. We show that if G satisfies P, then G has the following two types: (1) G is isoclinic to some stem groups of order p5, which form an isoclinic family. (2) G is isoclinic to a special p-group of exponent p. Elementary structures of groups with P are determined.


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## Body

## 1. Preliminaries

Letp be a prime and G be a finite p-group. A group G is said to have the property P, if any abelian subgroup M of G satisfies |MZ(G)/Z(G)|≤p. It is well known that if MZ(G)/Z(G) is cyclic, then M is abelian. It rises to consider the groups with property P.Mann [1] obtained the structure of 2-groups with property P; leting any abelian subgroup M of 2-group G satisfy |MZ(G)/Z(G)|≤p, then G is isoclinic to a dihedral group. And for any odd prime p he showed that if any abelian subgroup M of G satisfies |MZ(G)/Z(G)|≤p, then G/Z(G) is an elementary abelian p-group or a nonabelian group of order p3 and exponent p.Lemma 1 (see [1]).
LetG be a finite p-group which satisfies the property P. Then G/Z(G) is an elementary abelian p-group or a nonabelian group of order p3 and exponent p.This paper further discusses groups with propertyP. In the following all considered groups are finite p-groups among which p is an odd prime.First we state some notions and lemmas.Definition 2.
GroupsG and H are said to be isoclinic, if there exist isomorphisms σ:G/Z(G)→H/Z(H) and θ:G′→H′ which are compatible, in the sense that (aZ(G))σ=cZ(H) and (bZ(G))σ=dZ(H) (c,d∈H-Z(H)) imply that [a,b]θ=[c,d].Definition 3.
A finitep-group G is called a stem group, if Z(G)≤G′.Definition 4.
GroupG is called a special p-group, if G is an elementary abelian p-group or G′=Z(G)=Φ(G) is an elementary abelian p-group.Lemma 5 (see [2]).
Finitep-groups are isoclinic to stem groups.Lemma 6 (see [3]).
LetG be a finite regular p-group, s≥0; then for any a,b∈G, aps=bps if and only if (a-1b)ps=1.Lemma 7 (see [4]).
LetG be a regular p-group; then for any i, Ωi(G) is the set of elements of order pi.


## 2. Main Results

We need the following result.Lemma 8.
Suppose thatG and H are isoclinic. If G satisfies P, then so does H.Proof.
Assume that isomorphismsσ:G/Z(G)→H/Z(H) and θ:G′→H′ are compatible. And assume that M is an abelian subgroup of H. Then there exists a subgroup N of G such that (N/Z(G))σ=MZ(H)/Z(H). Suppose that c,d∈N∖Z(G) and (cZ(G))σ=aZ(H),(dZ(G))σ=bZ(H). Then a,b∈M∖Z(H). Then [c,d]θ=[a,b]=1 since σ and θ are compatible. It follows that |N/Z(G)|⩽p since G satisfies P. Hence |MZ(H)/Z(H)|⩽p. This completes the proof.Now we state our main result.Theorem 9.
LetG be a finite p-group satisfying property P, then
(1)
Gis isoclinic to some stem groups of order p5, which form an isoclinic family, or
(2)
G is isoclinic to a special p-group of exponent p.Proof.
By Lemma1, G/Z(G) is an elementary abelian p-group or a nonabelian group of order p3 and exponent p.
(1) Suppose thatG/Z(G) is not abelian.
ThenG/Z(G) is of order p3 and exponent p. By Lemma 5 we may suppose that Z(G)⩽G′. Noting that the nilpotent class of G is 3, thus Z(G)≠G′. It follows that G′ is abelian since |G′/Z(G)|=p. Assuming that G/Z(G)=〈aZ(G),bZ(G)〉, then G=〈a,b〉 since Z(G)<G′. Let H=〈a,ab〉=〈a,[a,b]〉. Then H is nilpotent of class 2 and thus H is a regular p-group. Note that ap∈Z(G) since G/Z(G) is of exponent p. So (b-1ab)p=b-1apb=ap, following that [a,b]p=(a-1b-1ab)p=1 by Lemma 6. Hence G′=〈[a,b]〉G=〈[a,b],[a,b,a],[a,b,b]〉 is abelian and |G′|⩽p3.If|G′|=p, then G is inner-abelian and G/Z(G)≅Zp2, a contradiction.If|G′|=p2, there are two cases.
Case  1  ([a,b,a]=1 or [a,b,b]=1). If [a,b,a]=1, then 〈G′,a〉 is abelian and 〈G′,a〉/Z(G)≅Zp2, a contradiction. For [a,b,b]=1, then 〈G′,b〉 is abelian and 〈G′,b〉/Z(G)≅Zp2, a contradiction.
Case  2  ([a,b,a]=[a,b,b]). Then we have [a,b,a][a,b,b-1]=[a,b,ab-1]=1. Thus H=〈[a,b],ab-1,Z(G)〉 is abelian and H/Z(G)≅Zp2, a contradiction. Hence |G′|=p3 and then |G|=p5.
Now we show that all groups of orderp5 with P are isoclinic.
Suppose thatG=〈a1,b1〉, H=〈a2,b2〉 are groups of order p5 with P. Then G/Z(G)=〈a1Z(G),b1Z(G)〉≅Ep3 and H/Z(H)=〈a2Z(H),b2Z(H)〉≅Ep3.
Setc1=[a1,b1], c2=[a2,b2] and define σ:G/Z(G)→H/Z(H) such that (a1Z(G))σ=a2Z(H), (b1Z(G))σ=b2Z(H).
Thenσ spans an isomorphism and (c1Z(G))σ=[a1Z(G),b1Z(G)]σ=[a2Z(H),b2Z(H)]=c2Z(H). Note that Z(G)=〈[c1,a1]〉×〈[c1,b1]〉,Z(H)=〈[c2,a2]〉×〈[c2,b2]〉 and G′=〈c1〉×Z(G), H′=〈c2〉×Z(H).
Settingc1θ=c2, [c1,a1]θ=[c2,a2], and [c1,b1]θ=[c2,b2], then θ deduces an isomorphism from G′ onto H′. Now we show that σ and θ are compatible.
For anyx,y∈G, write x=a1i1b1i2c1i3z1, y=a1j1b1j2c1j3z2, where z1,z2∈Z(G). Then
(1)(xZ(G))σ=(a1i1b1i2c1i3Z(G))σ(xZ(G))σ=(a1i1Z(G))σ(b1i2Z(G))σ(c1i3Z(G))σ(xZ(G))σ=a2i1Z(H)b2i2Z(H)c2i3Z(H)=a2i1b2i2c2i3Z(H).
Similarly,(yZ(G))σ=a2j1b2j2c2j3Z(H). Note that
(2)[x,y]=[a1i1b1i2c1i3z1,a1j1b1j2c1j3z2][x,y]=[a1i1b1i2c1i3,a1j1b1j2c1j3][x,y]=[a1i1b1i2,a1j1b1j2c1j3][c1i3,a1j1b1j2c1j3][x,y]=×[a1i1b1i2,a1j1b1j2c1j3,c1i3][x,y]=[a1i1b1i2,a1j1b1j2][a1i1,c1j3][b1i2,c1j3][c1i3,a1j1][x,y]=×[c1i3,b1j2][x,y]=[c1,a1]i3j1-i1j3[c1,b1]i3j2-i2j3[a1i1b1i2,a1j1b1j2][x,y]=[c1,a1]i3j1-i1j3[c1,b1]i3j2-i2j3[x,y]=×[a1i1b1i2,a1j1][a1i1b1i2,b1j2][a1i1b1i2,a1j1,b1j2][x,y]=[c1,a1]i3j1-i1j3[c1,b1]i3j2-i2j3[c1,a1](i12)(i21)[x,y]=×[c1,b1](i11)(i22)[x,y]=×[c1,a1](i21)(j12)[c1,b1](j11)(i22)[x,y]=×[c1,b1]i1i2j2-i2j1j2c1i1j2-i2j1[x,y]=[c1,a1]i3j1-i1j3+(i21)(j12)+(i12)(j21)[x,y]=×[c1,b1]i3j2-i2j3+i1i2j2-i2j1j2+(j11)(i22)+(i11)(j22)c1i1j2-i2j1.
Set(3)f1=i3j1-i1j3+(i21)(j12)+(i12)(j21),f2=i3j2-i2j3+i1i2j2-i2j1j2+(j11)(i22)+(i11)(j22),f3=i1j2-i2j1.
Then[x,y]=c1f3[c1,a1]f1[c1,b1]f2. Thus [x,y]θ=c2f3[c2,a2]f1[c2,b2]f2=[a2i1b2i2c2i3,a2j1b2j2c2j3]. Hence σ and θ are compatible.
(2)   Suppose thatG/Z(G) is an elementary abelian p-group.
Note thatG′≤Z(G) since G/Z(G) is abelian. By Lemma 5 we may suppose that Z(G)=G′. If expG=p, then G is a special p-group of exponent p. So suppose that expG>p.
LetG/Z(G)=〈a¯1〉×〈a¯2〉×⋯×〈a¯n〉, where a¯i=aiZ(G). Then G=〈a1,a2,…,an〉 and aip∈Z(G). Note that G′=〈[ak,al]〉G and [ak,al]p=[akp,al]=1. Hence G′ is an elementary abelian p-group of exponent p. Suppose that G′=〈ckl∣fm(ckl)=1〉, where ckl=[ak,al] and fm(ckl)=1 indicate the laws of G′. Suppose that K=〈dkl∣fm(dkl)=1〉 and K≅G′. By Schreier group expansion theories, we can add elements b1,b2,…,bn of order p such that [bk,bl]=dkl into group K, note that H={b1,b2,…,bn∣bip=1,[bk,bl]=dkl}. Then H is a group by Schreier group expansion theories.
Define a mapσ:ckl↦dkl for k,l=1,…,n. Then σ deduces an isomorphism from G′ onto H′. Note that H/H′=〈b¯1〉×〈b¯2〉×⋯×〈b¯n〉 and G/G^′=〈a¯1〉×〈a¯2〉×⋯×〈a¯n〉.
Settingσ1:aiG′↦biH′, then σ1 spans an isomorphism from G/G′ onto H/H′.
For anyxG′,yG′∈G/G′, assume that x=a1i1a2i2⋯aninc1,y=a1j1a2j2⋯anjnc2, where c1,c2∈G′.
Then(xG′)σ1=(a1i1a2i2⋯aninG′)σ1=b1i1b2i2⋯bninH′(yG′)σ1=(a1j1a2j2⋯anjnG′)σ1=b1j1b2j2⋯bnjnH′. So [x,y]σ2=[a1i1a2i2⋯aninc1,a1j1a2j2⋯anjnc2]σ2=(∏1⩽k⩽l⩽n‍[ak,al]ikjl-iljk)σ2=∏1⩽k⩽l⩽n‍[bk,bl]ikjl-iljk=[b1i1b2i2⋯bnin,b1j1b2j2⋯bnjn].
Thusσ1 and σ2 are compatible.
For anyx∈Z(H), write x=b1i1b2i2⋯bninc, where c∈H′. Then [b1i1b2i2⋯bninc,bi]=1.
Note that(b1i1b2i2⋯bninH′)σ1-1=(b¯1i1b¯2i2⋯b¯nin)σ1-1=a¯1i1a¯2i2⋯a¯nin=a1i1a2i2⋯aninG′, and similarly (biH′)σ1-1=aiG′. Then [b1i1b2i2⋯bnin,bi]σ1-1=[a1i1a2i2⋯anin,ai]=1. Hence a1i1a2i2⋯anin∈Z(G)=G′.
It follows thatb1i1b2i2⋯bninH′=(G′)σ1=H′ and thus b1i1b2i2⋯bnin∈H′. So Z(H)⩽H′. As a result H′=Z(H). Since H={b1,b2,…,bn∣bip=1,[bk,bl]=dkl}. We know that ℧1(H)=〈b1p,b2p,…,bnp〉=1; then expH=p. Hence G is isoclinic to a special p-group of exponent p.Note. In the sense of isoclinism, to investigate groups with P is to consider the special p-groups of exponent p.We need the following result.Lemma 10.
LetG be a special p-group of exponent p. Then(1)
ifd(G)=n, then |G′|≤pn(n-1)/2;(2)
for anyx,y∈G, if 〈xZ(G)〉=〈yZ(G)〉, then G satisfies P.Then we deduce the following.Theorem 11.
LetG be a special p-group of exponent p and d(G)=n, then(1)
if|G′|<pn-1, then G does not satisfy P.(2)
if|G′|=pn(n-1)/2, then G satisfies P.Proof.
Suppose thatG=〈a1,a2,…,an〉. Since G′ is an elementary abelian p-group, we may see G′ as an additive group of the vector space on GF(p).
(1) Assume that|G′|<pn-1. Then [a1,a2],[a1,a3],…,[a1,an] are linearly dependent. So there exist some integers which are not all 0 mod p such that [a1,a2]k1[a1,a3]k2⋯[a1,an]kn-1=1. Thus 〈a1,a2k1a3k2⋯ankn-1〉 is abelian. However, note that |〈a1,a2k1a3k2⋯ankn-1〉Z(G)/Z(G)|≠p since a1,a2,…,an are generators of G. So G does not have P.
(2) Assume that|G′|=pn(n-1)/2. Then G′=〈[a1,a2],…,[a1,an],…,[an-1,an]〉. For any x,y∈G, assume that x=a1i1a2i2⋯aninc1 and y=a1j1a2j2⋯anjnc2, where c1,c2∈G′. Then [x,y]=[a1i1a2i2⋯aninc1,a1j1a2j2⋯anjnc2]=[a1i1a2i2⋯anin,a1j1a2j2⋯anjn]=∏1⩽k⩽l⩽n‍[ak,al]ikjl-iljk.
Looking atG′ as an additive group of the vector space on GF(p), then [a1,a2],…,[a1,an],…,[an-1,an] are linearly independent. If [x,y]=1, then ikjl-iljk=0mod p for 1⩽k⩽l⩽n. Thus il/jl=i2/j2⋯=in/jn. Setting il/jl=m, then x=a1j1ma2j2m⋯anjnmc1=(a1j1a2j2)mc2(m2)a3j3m⋯anjnmc1=(a1j1a2j2,…,anjnc2)mc=ymc, where c∈G′. Hence 〈xZ(G)〉=〈ymZ(G)〉⩽〈yZ(G)〉. By Lemma 10  G satisfies the property P.Corollary 12.
LetG be a special p-group of exponent p. Then(1)
ifd(G)=2, then G satisfies the property P.(2)
ifd(G)=3, then G satisfies the property P if and only if |G′|=p3.Proof.
(1) Ifd(G)=2, then G is an inner-abelian p-group. Obviously, G has the property P.
(2) Assume thatd(G)=3 and G=〈a1,a2,a3〉. By Theorem 11   G satisfies P if |G′|=p3, but it does not if |G′|=p. Assuming that |G′|=p2, then [a1,a2], [a1,a3], [a2,a3] are linearly dependent. So there exist integers k1,k2,k3 which are not all 0 mod p such that [a1,a2]k1[a1,a3]k2[a2,a3]k3=1. Hence 〈a1,a2k1a3k2a2ik3a3jk3〉 is abelian.
Sincea1, a2, and a3 are generators of G, |〈a1,a2k1a3k2a2ik3a3jk3〉Z(G)/Z(G)|≠p, then G does not satisfy P;Theorem 13.
Assume thatG=〈a1,…,an〉 (n≥4) is a special p-group of exponent p and |G′|=pm, where n-1≤m<n(n-1)/2. Suppose that G′=〈[ak1,al1],…,[akn-1,aln-1],…,[akm,alm]〉 for ki,li∈{1,2,…,n}. If G satisfies the following properties, then G has P.(1)
[ak,al]≠1 for any k≠l.(2)
For any distinctl≠m∈{1,…,n} and for not all 0 mod  p integers i,j,[ak,al]i≠[ak,am]j.(3)
For[ak1,al1],[ak2,al2],…,[akn-1,aln-1], which satisfy {k1,l1,k2,l2,…,kn-1,ln-1}={1,2,…,n} and ki+1∈{ki,li} or li+1∈{ki,li}, where i=1,2,…,n-2.(4)
Any[aki,ali] of [akm+1,alm+1],…,[akn(n-1)/2,aln(n-1)/2] can be expressed as [aki,ali]=[akn,aln]t1i[akn+1,aln+1]t2i⋯[akm,alm]t(m-n+1)i, where t1i,t2i,…,t(m-n+1)i are integers.Proof.
For anyx,y∈G, write x=a1i1a2i2⋯aninc1 and y=a1j1a2j2⋯anjnc2, where c1,c2∈G′. Then
(4)[x,y]=[a1i1a2i2⋯aninc1,a1j1a2j2⋯anjnc2][x,y]=[ak1,al1]ik1jl1-il1jk1[ak2,al2]ik2jl2-il2jk2⋯[akn-1,aln-1]f1[x,y]=×[akn,aln]f2⋯[akm,alm]f3,
where(5)f1=ikn-1jln-1-iln-1jkn-1,f2=iknjln-ilnjkn+∑s=m+1n(n-1)/2‍t1s(iksjls-ilsjks),f3=ikmjlm-ilmjkm+∑s=m+1n(n-1)/2‍t(n-m+1)s(iksjls-ilsjks).
Note that[ak1,al1],[ak2,al2],…,[akn-1,aln-1],[akn,aln],…,[akm,alm] are linearly independent.
Hence if[x,y]=1 then ikrjlr-ilrjkr=ikrjlr-ilrjkr+∑s=m+1n(n-1)/2‍trs(iksjls-ilsjks)=0 mod p, where r=1,2,…,n,…,m. It follows that il/jl=i2/j2⋯=in/jn by (1) and (2). Supposing that il/jl=m, then x=ymc and thus 〈xZ(G)〉=〈ymZ(G)〉⩽〈yZ(G)〉. Hence |〈x,y,Z(G)〉/Z(G)|=|〈y,Z(G)〉/Z(G)|=p, following that G has P.Corollary 14.
LetG be a special p-group of exponent p and d(G)=n (n≥4). If |G′|=pn-1 or pn, then G does not satisfy the property P.Proof.
By Theorem13 if |G′|=pn-1 or pn, then G does not satisfy the conditions of Theorem 9. So G does not have P.



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