1 Introduction
All groups considered in this paper are finite.
Let G be a group admitting an action of a group A and let
CG(A)={x∈G∣xa=x for all a∈A}
be the fixed-point group of A in G. Thompson proved in [8] that if A is cyclic of prime order and CG(A)=1, then G must be nilpotent. In general it seems interesting to investigate how the structure of CG(A) influences the structure of G.
Throughout this paper, r is a fixed prime, A is an elementary
abelian r-group and G is an r′-group. We write A# for the set of non-identity elements of A.
From Glauberman’s solvable signalizer functor theorem [3] it follows that G must be solvable if |A|⩾r3 and CG(a) is solvable for each a∈A#. Moreover, it was proved in [4] using the classification of the finite simple groups that the above result is still true when |A|⩾r2. On the other hand, Ward [9] showed that if |A|⩾r3 and CG(a) is nilpotent for each a∈A#, then G is nilpotent. Ward also proved in [10] that if |A|⩾r4 and CG(a)′ is nilpotent for each a∈A#, then the derived group G′ of G is nilpotent.
Recall a group G is called supersolvable if every chief factor of G is cyclic. In this paper, we mainly focus on the following question.
Question.
What can be said about the structure of G if CG(a) is supersolvable for each a∈A#?
We shall prove the following result.
Theorem 1.1.
Let A be an elementary abelian r-group of order at least r4 acting on an r′-group G.
Suppose that CG(a) is supersolvable for each a∈A#. Then G is supersolvable.
The following example essentially due to Ward [10] shows that the conclusion of Theorem 1.1 may fail if |A|=r3. Let r,q,s,p be distinct primes such that
r∣q-1,r,q∣s-1 and r,q,s∣p-1;
it is easy to obtain such primes using Dirichlet’s theorem on primes in arithmetic progression. Let α1,α2,α3 be primitive r-th roots of unity in the fields 𝔽q,𝔽s,𝔽p, respectively.
Let U2=W1⋊〈ϕ1〉, where W1=〈x〉, o(x)=q, o(ϕ1)=r and xϕ1=xα1, and consider the regular wreath product V2=Cs≀regU2 (see [1, Chapter A, Definition 18.7]), where Cs is the cyclic group of order s.
Let W2 be the normal Sylow s-subgroup of V2; clearly V2=W2U2. Define an automorphism ϕ2 of V2 by
wu↦wα2u for w∈W2,u∈U2.
Similarly,
let U3=W2⋊〈ϕ2〉, and set V3=Cp≀regU3. Let W3 be the normal Sylow p-subgroup of V3 and let ϕ3 be an automorphism of V3 given by
wu↦wα3u for w∈W3,u∈U3.
Let
H=V3⋊〈ϕ3〉=W3W2W1〈ϕ1〉〈ϕ2〉〈ϕ3〉
and take G=W3W2W1 and A=〈ϕ1〉〈ϕ2〉〈ϕ3〉. Observe that G is an r′-group and A is elementary abelian of order r3. The action of A on G
is defined by conjugation in H. For each a=ϕ1iϕ2jϕ3k∈A#, where i,j,k∈{0,1,…,r-1}, Lemma 2.1 (d) below gives
CG(a)={CW3(a)CW2(a)if i≠0,CW3(a)W1if i=0 and j≠0,W2W1if i=j=0 and k≠0.
It is easy to see from Lemma 3.1 below that CG(a) is supersolvable and it is also easy to show that G′ is not nilpotent. Hence G is not supersolvable.
Denote by 𝐅(G) the Fitting subgroup of a group G and denote by 𝐅i(G) the inverse image of 𝐅(G/𝐅i-1(G)) in G for i⩾2.
Theorem 1.2.
Let A be an elementary abelian r-group of order at least r3 acting on an r′-group G.
Suppose that CG(a) is supersolvable for each a∈A#. Then G′⩽F3(G). Furthermore, G′⩽F2(G) if one of the following statement holds:
CG(a)′ is of odd order for each a∈A#, or
r is not a half-Fermat prime.
Recall that a prime r is called half-Fermat if there exists a prime p such that r=12(pm+1) for some m∈ℕ. In the proof of Theorems 1.1 and 1.2, the following theorem plays an important role.
Theorem 1.3.
Let A be an elementary abelian r-group of order at least r3 acting on an r′-group G.
If p is a prime and CG(a) is p-nilpotent for each a∈A#, then G is p-nilpotent.
2 Proof of Theorem 1.3
First we recall some basic results on coprime action.
Lemma 2.1.
Let G be a group admitting an action of a group A with (|G|,|A|)=1.
Then the following assertions hold.
Suppose that A is abelian and G is a non-trivial elementary abelian group. If CG(a)=1 for every a∈A#, then A is cyclic.
If G has the Sylow
π
-property for a set
π
of primes, then G has an A-invariant Hall
π
-subgroup.
If N⊴G and N is A-invariant, then CG/N(A)=CG(A)N/N.
If G=XY, where X and Y are A-invariant subgroups, then
CG(A)=CX(A)CY(A).
Suppose that A is non-cyclic. Then
G=〈CG(B)∣B⩽A and A/B is cyclic〉.
In particular, G=〈CG(a)∣a∈A#〉.
If H is an A-invariant Hall
π
-subgroup of G, then CH(A) is a Hall
π
-subgroup of CG(A).
Proof.
Assertions
(a)–(f) follow from [6, 8.3.2, 8.2.6 (a), 8.2.2, 8.2.11, 8.3.4 (a,b), 8.2.6 (e)], respectively.
∎
Lemma 2.2 ([2, Theorem 2.3]).
Let F be a field, G a group and V an F[G]-module. Suppose that H⩽G, char(F) does not divide |G:H| and
V is a completely reducible F[H]-module. Then V is a completely reducible F[G]-module.
Proof of Theorem 1.3.
Suppose the theorem false and let G be a counter-example of minimal order. By the properties of coprime action and the minimality of G, all proper A-invariant subgroups of G and all quotient groups G/N modulo non-trivial A-invariant normal subgroups N of G are p-nilpotent. We complete the proof in three steps.
Step 1: We claim that G is p-solvable and Op′(G)=1.
If p=2, then CG(a) is solvable and by [6, Theorem 11.3.3], G is solvable, as desired.
Now assume that p is odd. By Lemma 2.1 (b), there exists an A-invariant Sylow p-subgroup P of G. If P=1, then clearly G is p-solvable so we may assume that P≠1; thus Z(J(P))≠1. Observe that Z(J(P)) and NG(Z(J(P))) are both A-invariant.
If NG(Z(J(P)))<G, then NG(Z(J(P))) is p-nilpotent by the minimality of G, and then G is p-nilpotent by Glauberman’s ZJ-theorem [5, Chapter X, Theorem 9.10], which is a contradiction. Hence NG(Z(J(P)))=G and it follows that G/Z(J(P)) is p-nilpotent, so that G is p-solvable.
If Op′(G)≠1, then by the minimality of G, G/Op′(G) is p-nilpotent, which implies that G is p-nilpotent, a contradiction.
Step 2: We claim that G is p-closed.
Observe that Op(G)≠1 by Step 1. It follows from the minimality of G that G/Op(G) is p-nilpotent. Let L/Op(G) be the Hall p′-subgroup of G/Op(G); thus G/L is a p-group and L is A-invariant. If L<G, then L is p-nilpotent. For the normal p-complement Lp′ of L we have Lp′⩽Op′(G)=1. Thus L is a p-subgroup and so is G, which is a contradiction. Hence G=L is p-closed.
Step 3: Final contradiction.
Write P=Op(G). As CG(P)⊴G, we have Op′(CG(P))⩽Op′(G)=1. Hence CG(P)=Z(P). Observe that
G=〈CG(B)∣B⩽A and A/B is cyclic〉
by Lemma 2.1 (e). Since G is not p-nilpotent, there exists a subgroup
B of A such that CG(B)⩽̸P. By hypothesis CG(B) has a normal p-complement K≠1. Now K⩽CG(B)⩽CG(b) for all b∈B#. Since G is p-closed, each CG(b) is p-closed and has a normal p-complement. As
K⩽CG(b), K centralizes CP(b). This means that K centralizes 〈CP(b)∣b∈B#〉. As |B|=r2, we have that K centralizes P=〈CP(b)∣b∈B#〉 and
this contradicts CG(P)=Z(P). This is the final contradiction.
∎
Corollary 2.3 ([9]).
Let A be an elementary abelian r-group of order at least r3 acting on an r′-group G.
If CG(a) is nilpotent for each a∈A#, then G is nilpotent.
Let ⪯ be a total order relation on the set ℙ of all primes. Recall that a group G is said to satisfy the Sylow tower property with respect to ⪯ if G has a normal Hall πp-subgroup for each prime p, where πp={q∣p⪯q}. If ⪯ is the natural order relation on ℙ, then G is said to satisfy the Sylow tower property of supersolvable type.
Corollary 2.4.
Let A be an elementary abelian r-group of order at least r3 acting on an r′-group G. Let ⪯ be a total order relation on P.
Suppose that CG(a) satisfies the Sylow tower property with respect to ⪯ for each a∈A#. Then G satisfies the Sylow tower property with respect to ⪯.
Proof.
We work by induction on |G|. Let p be the smallest prime with respect to ⪯ dividing the order of G. Observe that CG(a) is p-nilpotent for each a∈A#. It follows from Theorem 1.3 that G is p-nilpotent. The normal Hall p′-subgroup H of G is clearly A-invariant. For each a∈A#, CH(a) is a subgroup of CG(a) satisfying the Sylow tower property with respect to ⪯. Observe that H<G since p divides the order of G. By induction, H satisfies the Sylow tower property with respect to ⪯. Thus G satisfies the Sylow tower property with respect to ⪯.
∎
Corollary 2.5.
Let A be an elementary abelian r-group of order at least r3 acting on an r′-group G. Suppose that n is a positive integer.
If CG(a) is abelian of exponent dividing n for each a∈A#, then G is abelian of exponent dividing n. In particular, if CG(a) is abelian for each a∈A#, then G is abelian.
Proof.
We have G=〈CG(B)∣B⩽A and A/B is cyclic〉 by Lemma 2.1 (e).
By hypothesis CG(B) is abelian of exponent dividing n for all subgroups B of A with A/B cyclic. Suppose that
B1 and B2 are maximal subgroups of A. Then, as |A|=r3, there exists b∈A# such that b∈B1∩B2. Hence CG(B1)CG(B2)⩽CG(b), which is abelian by hypothesis. Therefore CG(B1)
and CG(B2) commute and it follows that G is abelian of exponent dividing n.
∎
3 Proof of Theorem 1.1
The following is a reformulation of a supersolvability criterion due to Reinhold Baer.
Lemma 3.1.
The group G is supersolvable if and only if
G satisfies the Sylow tower property of supersolvable type and
NG(Gp)/CG(Gp)Gp is abelian of exponent dividing p-1 for every prime p and every Sylow p-subgroup Gp of G.
Proof of Theorem 1.1.
Suppose Theorem 1.1 is false. Let G be a counter-example of minimal order. By the minimality of G, all proper A-invariant subgroups of G and all quotient groups G/N modulo non-trivial A-invariant normal subgroups N of G are supersolvable.
Let p be the largest prime dividing |G| and let P be a Sylow p-subgroup of G. It follows from Corollary 2.4 that G satisfies the Sylow tower property of supersolvable type, and so P⊴G and P⩽𝐅(G). The rest of the proof is divided into three steps:
Step 1: We claim that Φ(G)=1 and P=𝐅(G) is the unique minimal A-invariant normal subgroup of G.
If Φ(G)≠1, then G/Φ(G) is supersolvable by the minimality of G. Hence G is supersolvable, which is a contradiction. Let
𝔛={1<N⊴G∣NA=N}.
Suppose that there exist minimal elements N1,N2∈𝔛 such that N1≠N2. Then N1∩N2=1 and since both G/N1 and G/N2 are supersolvable, so is
G=G/(N1∩N2), contrary to the choice of G. Therefore 𝔛 has a unique minimal member, and 𝐅(G)=P is a p-group since P⩽𝐅(G).
Since Φ(P)⩽Φ(G)=1, P is an elementary abelian p-group, and it can be viewed as a 𝔽p[AG]-module.
By [7, Theorem 1.12], P=𝐅(G) is a completely reducible 𝔽p[G]-module; since (|A|,|G|)=1, it follows from Lemma 2.2 that P is a completely reducible 𝔽p[AG]-module. However P∈𝔛 and since 𝔛 has a unique minimal member, P must be an irreducible 𝔽p[AG]-module, that is, P is the unique minimal A-invariant normal subgroup of G, as claimed.
Step 2: Let H be the A-invariant Hall p′-subgroup of G. We claim that every A-invariant proper subgroup of H is abelian of exponent dividing p-1.
Let X be an A-invariant proper subgroup of H. Then Y=PX is an A-invariant proper subgroup of G. By the minimality of G, Y is supersolvable. It follows from Lemma 3.1 that Y/PCY(P)=NY(P)/PCY(P) is abelian of exponent dividing p-1. Observe that CG(P)⩽P since P=𝐅(G). Hence CY(P)⩽P and thus CY(P)=P since P is abelian. Then we have that X≅Y/P is abelian of exponent dividing p-1, as claimed.
Step 3: Final contradiction.
Let B=CA(H). First assume that |B|⩾r2. For each b∈B#,
we have the identity
CG(b)=CP(b)H since H⩽CG(b). This implies that CP(b) is normalized by H since CG(b) is supersolvable. Observe that P is abelian. Immediately we have that CP(b)⊴G. It is easy to see that CP(b) is an A-invariant normal subgroup of G. By Step 1, P is the minimal A-invariant normal subgroup of G. Thus CP(b)=1 or CP(b)=P for each b∈B#.
It follows from Lemma 2.1 (a) that
CP(b1)=P for some b1∈B#. This implies that G=PH⩽CG(b1) is supersolvable, which is a contradiction. Hence |B|⩽r.
For each b∈A∖B, we have that CH(b) is an A-invariant proper subgroup of H. By Step 2, CH(b) is abelian of exponent dividing p-1. Considering the faithful action of A/B on H, we see that CH(bB)=CH(b) is abelian of exponent dividing p-1 for each bB∈(A/B)#. Since |A/B|⩾r3, it follows from Corollary 2.5 that H is abelian of exponent dividing p-1. It now follows easily from Lemma 3.1 that
G is supersolvable. This contradiction completes the proof.
∎
4 Proof of Theorem 1.2
The following lemma is a special case of [2, Theorem A].
Lemma 4.1.
Let R be a group of order r that acts on the solvable r′-group H. Let V be an RH-module and a faithful, completely reducible H-module.
Suppose that K is a nilpotent normal subgroup of CH(R) such that
K⩽Ker(CH(R) on CV(R)).
Then K⩽F2(H). Furthermore, if r is not a half-Fermat prime or K is of odd order, then K⩽F(H).
Proof.
Let L be the subnormal closure of K in H. By [2, Theorem A] we have L=K[L,R] and [L,R] is nilpotent.
Since L/[L,R] is isomorphic to a quotient of K and K is nilpotent, L/[L,R] is nilpotent. Observe that L=𝐅2(L)⩽𝐅2(H) since L is subnormal in H. Hence K⩽L⩽𝐅2(H).
Write [L,R]=S×P with S a 2-group and P a 2′-group. Using [2, Theorem A] again,
we conclude from the nilpotency of K that S=1 and we also find that P=1 if r is not a half-Fermat prime or K is of odd order. Hence in these cases, [L,R]=1 and K=L is subnormal in H and so K⩽𝐅(H).
∎
For the proof of Theorem 1.2, we need the following two results.
Theorem 4.2.
Let A be an elementary abelian r-group of order at least r3 acting on an r′-group G.
Suppose that CG(a) is supersolvable for each a∈A#. Then G′⩽F3(G).
Proof.
Suppose that the theorem is false and let G be a counter-example of minimal order. Thus all proper A-invariant subgroups M of G satisfy M′⩽𝐅3(M), and all quotient groups G/N with N a non-trivial A-invariant normal subgroup N of G satisfy (G/N)′⩽𝐅3(G/N).
By Corollary 2.4, G satisfies the Sylow tower property of supersolvable type. Let p be the largest prime dividing |G| and let V be the Sylow p-subgroup of G. Clearly V is an A-invariant normal subgroup of G and V⩽𝐅(G).
Step 1: We claim that Φ(G)=1 and V=𝐅(G) is an irreducible 𝔽p[AG]-module.
Suppose that Φ(G)≠1. Considering the quotient group G/Φ(G), we conclude from the minimality of G that (G/Φ(G))′⩽𝐅3(G/Φ(G))=𝐅3(G)/Φ(G). This implies that G′⩽𝐅3(G), contrary to the choice of G. Hence 𝐅(G) can be viewed as an AG-module, possibly of mixed characteristic.
By [7, Theorem 1.12] and Lemma 2.2, 𝐅(G) is a completely reducible AG-module. If the AG-module 𝐅(G) is reducible, then we may assume 𝐅(G)=V1×V2, where 1≠Vi is an AG-module for each i. Consider the group G/Vi and let Ti/Vi=𝐅3(G/Vi). By the minimality of G we have G′⩽G′Vi⩽Ti. Observe that
T1∩T2≲(T1∩T2)V1/V1×(T1∩T2)V2/V2
is a subgroup of T1/V1×T2/V2 whose Fitting height is at most 3.
We have T1∩T2=𝐅3(T1∩T2)⩽𝐅3(G) since
T1∩T2⊴G.
Observe that 𝐅3(G)⩽T1∩T2. Thus T1∩T2=𝐅3(G) and so G′⩽T1∩T2=𝐅3(G), which is a contradiction. Hence 𝐅(G) is an irreducible 𝔽p[AG]-module. Since 1≠V⩽𝐅(G) and V is an A-invariant normal subgroup of G, we have V=𝐅(G), as claimed.
Step 2: We claim that 𝐅(CG(a))⩽𝐅3(G) for each a∈A#.
Let H be the A-invariant Hall p′-subgroup of G. For each a∈A#, we see from Lemma 2.1 (f) that CV(a)∈Sylp(CG(a)) and CH(a)∈Hallp′(CG(a)).
Since V⊴G, we have CV(a)⊴CG(a).
Clearly CV(a)⩽𝐅(CG(a)) and CV(a)∈Sylp(𝐅(CG(a))). Let K be the Hall p′-subgroup of 𝐅(CG(a)). Then K is a nilpotent normal subgroup of CH(a) such that
K⩽Ker(CH(a) on CV(a)).
By Step 1, V is an irreducible AG-module. It is easy to see that V is an irreducible AH-module. Thus V is a completely reducible H-module and H acts faithfully on V since CG(V)⩽V. Applying Lemma 4.1, we have K⩽𝐅2(H). Thus 𝐅(CG(a))=CV(a)K⩽V𝐅2(H)⩽𝐅3(G), as claimed.
Step 3: Final contradiction.
Considering the action of A on G/𝐅3(G), by Lemma 2.1 (c) we see that
CG/𝐅3(G)(a)=CG(a)𝐅3(G)/𝐅3(G)≅CG(a)/C𝐅3(G)(a)
is a homomorphic image of CG(a)/𝐅(CG(a)) since
𝐅(CG(a))⩽𝐅3(G) by Step 2.
Observe that CG(a)/𝐅(CG(a)) is abelian since CG(a) is supersolvable. Hence CG/𝐅3(G)(a) is abelian for each a∈A#. Since |A|⩾r3, by Corollary 2.5 the group G/𝐅3(G) is abelian and so G′⩽𝐅3(G). This is a contradiction and the proof is complete.
∎
Theorem 4.3.
Let A be an elementary abelian r-group of order at least r3 acting on an r′-group G.
Suppose that CG(a) is supersolvable for each a∈A#. If either
CG(a)′ has odd order for each a∈A#, or
r is not a half-Fermat prime,
then G′⩽F2(G).
Proof.
Suppose the theorem false and let G be a counter-example of minimal order. Let p be the largest prime dividing |G|. Let V be the Sylow p-subgroup of G.
Arguing as in the proof of Theorem 4.2, we see that Φ(G)=1 and V=𝐅(G) is an irreducible 𝔽p[AG]-module.
Next we claim that CG(a)′⩽𝐅2(G) for each a∈A#.
Let H be the A-invariant Hall p′-subgroup of G. For each a∈A#, by Lemma 2.1, CV(a)∈Sylp(CG(a)) and CH(a)∈Hallp′(CG(a)). Since V⊴G, we have CV(a)⊴CG(a) and clearly CV(a)⩽𝐅(CG(a)).
Let K be
a Hall p′-subgroup of CG(a)′. Since CG(a)′ is nilpotent, K is a characteristic subgroup of CG(a)′. Hence K is a nilpotent normal p′-subgroup of CG(a), moreover, K⩽CH(a). Observe that
CV(a)⩽𝐅(CG(a)) and K⩽CG(a)′⩽𝐅(CG(a)). Hence [CV(a),K]=1, that is,
K⩽Ker(CH(a) on CV(a)).
Since V is an irreducible AG-module, we have V⩽Ker(AG on V). Let U be a non-zero AH-module of V.
Since U is V-invariant, U is AG-invariant, that is, U is an AG-submodule of V.
Thus V is also an irreducible AH-module, and therefore V is a completely reducible H-module and H acts faithfully on V since CG(V)⩽V.
By hypothesis, either |K| is odd or r is not a half-Fermat prime. Applying Lemma 4.1, we conclude that K⩽𝐅(H). Thus
CG(a)′⩽CV(a)K⩽V𝐅(H)⩽𝐅2(G),
as claimed.
Finally, considering the action of A on G/𝐅2(G), for each a∈A#, we see from Lemma 2.1 (c) that CG/𝐅2(G)(a) is abelian. Since |A|⩾r3, by Corollary 2.5, G/𝐅2(G) is abelian, that is, G′⩽𝐅2(G). This is a contradiction and the proof is complete.
∎
and
Xiuyun Guo
