Fixed point conditions for non-coprime actions

Abstract

In the setting of finite groups, suppose $J$ acts on $N$ via automorphisms so that the induced semidirect product $N\rtimes J$ acts on some non-empty set $\Omega$, with $N$ acting transitively. Glauberman proved that if the orders of $J$ and $N$ are coprime, then $J$ fixes a point in $\Omega$. We consider the non-coprime case and show that if $N$ is abelian and a Sylow $p$-subgroup of $J$ fixes a point in $\Omega$ for each prime $p$, then $J$ fixes a point in $\Omega$. We also show that if $N$ is nilpotent, $N\rtimes J$ is supersoluble, and a Sylow $p$-subgroup of $J$ fixes a point in $\Omega$ for each prime $p$, then $J$ fixes a point in $\Omega$.

1. Introduction

Suppose a finite group $J$ acts via automorphisms on a finite group $N$ and the induced semi-direct product $G= N \rtimes J$ acts on some non-empty set $\Omega$ where the action of $N$ is transitive. Glauberman showed that if each supplement $H$ of $N$ in $G$ splits over $N \cap H$ and each complement of $N$ in $G$ is conjugate to $J$, then there exists a $J$-invariant element $\omega \in \Omega$. Consequently, if the orders of $J$ and $N$ are coprime so that the Schur–Zassenhaus theorem applies, a fixed point always exists [4, Thm. 4]. In this note, we consider the non-coprime case and establish some conditions for the existence of a fixed point.

Given an action as described above, consider the stabiliser $G_\alpha \leq G$ fixing an arbitrary point $\alpha \in \Omega$. As $N$ is transitive, $G_\alpha$ supplements $N$ in $G$. In this context, $J$ fixes an element of $\Omega$ if and only if the following two conditions are met. Firstly, we must ensure $G_\alpha$ splits over $N\cap G_\alpha$ so that there exists some complement $J'$. As $G/N \cong G_\alpha /(N\cap G_\alpha )$, it will follow that $J'$ also complements $N$ in $G$. Secondly, we require that $J' = g^{-1}Jg$ for some $g\in G$ so that $J$ fixes $g\cdot \alpha$. For the latter requirement, we concern ourselves with conditions for two specific complements in a semidirect product to be conjugate.

To this end, we say two subgroups $H$ and $H'$ are locally conjugate in a group $G$ if for each prime $p$, a Sylow $p$-subgroup of $H$ is conjugate to a Sylow $p$-subgroup of $H'$. Losey and Stonehewer showed that if $H$ and $H'$ are locally conjugate supplements of some normal nilpotent subgroup $N$ in a soluble group $G$, then $H$ and $H'$ are conjugate if either $G/N$ is nilpotent or $N$ is abelian [7]. Evans and Shin further showed that if $N$ is abelian, then $G$ need not be soluble [3].

We first restrict $N$ to be abelian and use a decomposition result from group cohomology to provide an alternate proof of:

Lemma 1.1 (Evans and Shin)

In a finite group, two complements of a normal abelian subgroup are conjugate if and only if they are locally conjugate.

We use this, along with Gaschütz's result that a finite group $G$ splits over an abelian subgroup $N$ if and only if for each prime $p$, a Sylow $p$-subgroup $S$ of $G$ splits over $N\cap S$, to show:

Theorem 1.2 Given a finite group $J$ acting via automorphisms on a finite abelian group $N$, suppose the induced semidirect product $N\rtimes J$ acts on some non-empty set $\Omega$ where the action of $N$ is transitive. If for each prime $p$, a Sylow $p$-subgroup of $J$ fixes an element of $\Omega$, then there exists some $J$-invariant element $\omega \in \Omega$.

This had previously been shown using elementary arguments for the special case that $J$ is supersoluble [2, Cor. 2]. The theorem implies:

Corollary 1.3 Let $G$ be a finite split extension over an abelian subgroup $N$. If for each prime $p$ there is a Sylow $p$-subgroup $S$ of $G$ such that any two complements of $N\cap S$ in $S$ are conjugate, then any two complements of $N$ in $G$ are $G$-conjugate.

This extends a result of D. G. Higman [5, Cor. 2] that requires the complements of $N\cap S$ in $S$ to be conjugate within $S$.

We then consider nilpotent $N$ and supersoluble $N\rtimes J$. We adapt our approach for lemma 1.1 to nonabelian cohomology and demonstrate:

Lemma 1.4 In a finite supersoluble group, two complements of a normal nilpotent subgroup are conjugate if and only if they are locally conjugate.

With this, we then show:

Theorem 1.5 Given a finite group $J$ acting via automorphisms on a finite nilpotent group $N$, suppose the induced semidirect product $N\rtimes J$ is supersoluble and acts on some non-empty set $\Omega$ where the action of $N$ is transitive. If for each prime $p$, a Sylow $p$-subgroup of $J$ fixes an element of $\Omega$, then there exists some $J$-invariant element $\omega \in \Omega$.

The theorem also implies an analogue of corollary 1.3 that we state and prove in § 3.

1.1 Outline

We proceed as follows. In the remainder of this section, we introduce notation and some conventions from group cohomology. In the next section, we restrict $N$ to be abelian and prove theorem 1.2. We then restrict $N$ to be nilpotent and $N\rtimes J$ to be supersoluble in § 3 and prove theorem 1.5, before concluding in § 4.

1.2 Notation and conventions

All groups in this note are assumed finite. A subgroup $K\leq G$ supplements $N\vartriangleleft G$ if $G=NK$ and complements $N$ if it both supplements $N$ and the intersection $N\cap K$ is trivial. We denote conjugation by $g^\gamma = \gamma ^{-1}g\gamma$ for $g,\,\gamma \in G$ and otherwise let groups act from the left. For a prime $p$, we let $\operatorname {Syl}_p(G)$ denote the set of Sylow $p$-subgroups of a group $G$.

We rely on rudimentary notions from group cohomology that can be found in the texts of Brown [1] and Serre [8]. Given a group $J$ acting on a group $N$ via automorphisms, crossed homomorphisms or 1-cocycles are maps $\varphi : J \to N$ satisfying $\varphi (jj') = \varphi (j) \varphi (j')^{j^{-1}}$ for all $j,\,j' \in J$. Two such maps $\varphi$ and $\varphi '$ are cohomologous if there exists $n\in N$ such that $\varphi '(j) = n^{-1} \varphi (j) n^{j^{-1}}$ for all $j\in J$; in this case, we write $\varphi \sim \varphi '$. We take the first cohomology $H^1(J,\,N)$ to be the pointed set $Z^1(J,\,N)$ of crossed homomorphisms modulo this equivalence. The distinguished point corresponds to the equivalence class containing the map taking each element of $J$ to the identity of $N$. Our interest in this set stems primarily from the well-known bijective correspondence [8, Exer. 1 in §I.5.1] between it and the $N$-conjugacy classes of complements to $N$ in $N\rtimes J$. Specifically, for each $\varphi \in Z^1(J,\,N)$, the subgroup $F(\varphi ) = \{ \varphi (j)j \}_{j\in J}$ complements $N$ in $NJ$ and all such complements may be written in this way. Two crossed homomorphisms yield conjugate complements under $F$ if and only if they are cohomologous, so $F$ induces the desired correspondence.

For a subgroup $K\leq J$, we let $\varphi |_{K}$ denote the restriction of $\varphi \in Z^1(J,\,N)$ to $K$ and $\operatorname {res}^J_K: H^1(J,\,N) \to H^1(K,\,N)$ be the map induced in cohomology. For $\varphi \in Z^1(K,\,N)$ and $j\in J$, define $\varphi ^j(x) = \varphi (x^{j^{-1}})^{j}$. We call $\varphi$ $J$-invariant if $\operatorname {res}^K_{K\cap K^j} \varphi \sim \operatorname {res}^{K^j}_{K\cap K^j} \varphi ^j$ for all $j\in J$ and let $\operatorname {inv}_J H^1(K,\,N)$ denote the set of $J$-invariant elements in $H^1(K,\,N)$. For any $\varphi \in Z^1(J,\,N)$, we have $\varphi ^j(x) = n^{-1} \varphi (x) n^{x^{-1}}$ where $n= \varphi (j^{-1})$ so that $\varphi ^j \sim \varphi$. In particular, $\operatorname {res}^J_K H^1(J,\,N) \subseteq \operatorname {inv}_J H^1(K,\, N)$.

2. $N$ is abelian

In this section, we restrict $N$ to be abelian so that $H^1(J,\,N)$ takes the form of an abelian group. We first prove lemma 1.1 as stated in § 1.

Proof of lemma 1.1 Suppose we are given locally conjugate complements $J$ and $J'$ of a normal abelian subgroup $N$ in some group $G$. As any element $g\in G$ may be uniquely written $g=jn$ for $j\in J$ and $n\in N$, for each prime $p$ we have $J_p' = (J_p)^n$ for some $J_p\in \operatorname {Syl}_p(J)$, $J_p' \in \operatorname {Syl}_p(J')$, and $n\in N$. Let $\varphi '\in Z^1(J,\,N)$ denote the crossed homomorphism corresponding to $J'$. It suffices to show that $\varphi '\sim 1$, where $1\in Z^1(J,\,N)$ denotes the map taking each element of $J$ to the identity of $N$. Through the $p$-primary decomposition of $H^1(J,\,N)$, we have the isomorphism [1, §III.10]:

(2.1)\begin{equation} H^1(J, N) \cong \oplus_{p\in\mathcal{D}} \operatorname{inv}_J H^1(J_p, N) \end{equation}

where $\mathcal {D}$ is the set of prime divisors of $\left\lvert{J}\right\rvert$ and the $J_p$ are those given above. For every $p\in \mathcal {D}$, we see that $\varphi '|_{J_p}\sim 1|_{J_p}$ as $J_p$ and $J_p'$ are $N$-conjugate complements of $N$ in $NJ_p$. Thus, $\varphi '$ maps to the identity in each direct summand on the right-hand side of (2.1) and we may conclude $\varphi '\sim 1$ so that $J$ and $J'$ are conjugate.

We can now use the lemma and Gaschütz's theorem to prove theorem 1.2.

Proof of theorem 1.2 Given $J$, $N$, and $\Omega$ as described in the hypotheses of the theorem, let $G= N \rtimes J$ denote the induced semidirect product and consider the stabiliser subgroup $G_\alpha$ for some fixed $\alpha \in \Omega$. As $N$ acts transitively, any $g\in G$ may be written $g\cdot \alpha = n\cdot \alpha$ for some $n\in N$, so that $n^{-1}g \in G_\alpha$. Thus, $G=NG_\alpha$.

We claim $G_\alpha$ splits over $N\cap G_\alpha$. For any prime $p$, there exists by hypothesis some $n\in N$ and $P \in \operatorname {Syl}_p(J)$ such that $P^n \leq G_\alpha$. Let $L\in \operatorname {Syl}_p(N\cap G_\alpha )$. As $\left\lvert{G_\alpha }\right\rvert = \left\lvert{N\cap G_\alpha }\right\rvert[G:N]$, it follows that $S=LP^n \in \operatorname {Syl}_p(G_\alpha )$ so $P^n$ complements $S \cap N = L$ in $S$. As the choice of prime $p$ was arbitrary, we may apply Gaschütz's theorem to conclude that $G_\alpha$ splits over $N\cap G_\alpha$.

Let $J'$ complement $N\cap G_\alpha$ in $G_\alpha$. As $G/N \cong G_\alpha / (N \cap G_\alpha )$, it follows that $J'$ also complements $N$ in $G$. Lemma 1.1 then implies that $J' = J^g$ for some $g\in G$ so that $J$ fixes $\omega = g\cdot \alpha$.

Finally, we outline how corollary 1.3 follows from theorem 1.2.

Proof of corollary 1.2 Given a group $G$ satisfying the hypotheses of the corollary, suppose $J$ and $J'$ each complement $N$ in $G$. Then $G$ acts on the cosets $\Omega = G/J'$ in such a way that we may apply theorem 1.2 to infer that $J$ fixes $gJ'$ for some $g\in G$. Therefore, $J$ and $J'$ are conjugate. As the choice of complements was arbitrary, we may conclude.

3. $N$ is nilpotent and $N\rtimes J$ is supersoluble

In this section, we suppose that $N$ is nilpotent and $N \rtimes J$ is supersoluble. Consequently, $N$ decomposes as the direct sum $N\cong \oplus _{p\in \mathcal {D}} N_p$ over its characteristic Sylow $p$-subgroups $N_p$ where $\mathcal {D}$ denotes the set of prime divisors of $\lvert{N}\rvert$. Direct calculations show that the natural projections $N \to N_p$ induce an isomorphism of pointed sets

(3.1)\begin{equation} H^1(J,N) \cong \oplus_{p\in D} H^1(J,N_p). \end{equation}

To parse the components on the right-hand side of (3.1), we introduce the following:

Proposition 3.1 Suppose a group $J$ acts on a $p$-group $N$ via automorphisms, so that the induced semidirect product $N\rtimes J$ is supersoluble. Then $\operatorname {res}_{J_p}^J: H^1(J,\,N) \to \operatorname {inv}_J H^1(J_p,\,N)$ is an isomorphism for $J_p\in \operatorname {Syl}_p(J)$.

Proof. We induct on the order of $J$. If $J$ itself is a $p$-group, the conclusion is immediate. If $p$ is not a divisor of $\lvert{J}\rvert$, the lemma follows from the Schur–Zassenhaus theorem. Otherwise, let $Q\vartriangleleft J$ be a Sylow $q$-subgroup where $q$ is the largest prime divisor of $\lvert{J}\rvert$ [6, Exer. 3B.10] so that $J\cong Q \rtimes M$ for some Hall $q'$-subgroup $M\leq J$. Consider the inflation–restriction exact sequence [8, §I.5.8],

(3.2)\begin{equation} 1 \to H^1(J/Q, N^Q) \to H^1(J,N) \xrightarrow{\operatorname{res}^J_Q} H^1(Q,N)^{J/Q} \end{equation}

where $N^Q$ denotes the elements of $N$ fixed by $Q$.

If $q\ne p$, then $H^1(Q,\,N)$ is trivial so that $H^1(J,\,N) \cong H^1(M,\, N^Q)$. In the supersoluble group $NQ$, $Q$ is a Sylow $q$-subgroup for the largest prime divisor of $\lvert{NQ}\rvert$, so that $Q\vartriangleleft N Q$ and $N^Q = N$. Consequently, $H^1(J,\,N) \cong H^1(M,\, N)$. We claim that $\operatorname {res}^J_M$ affords this isomorphism. It suffices to show that $\operatorname {res}^J_M$ is surjective. For any $\varphi \in Z^1(M,\, N)$, we may define $\tilde \varphi : J \to N$ by $\tilde \varphi (qm) = \varphi (m)$ for $q\in Q$ and $m\in M$. This map is well-defined as $J\cong Q \rtimes M$. For $q,\,q'\in Q$ and $m,\,m'\in M$, we have $\tilde \varphi (qmq'm') = \varphi (mm') = \varphi (m)\varphi (m')^{m^{-1}} = \tilde \varphi (qm)\tilde \varphi (q'm')^{(qm)^{-1}}$, where the last equality follows from the fact that elements of $N$ commute with elements of $Q$. Thus, $\tilde \varphi \in Z^1(J,\, N)$. As $\tilde \varphi |_M = \varphi$, we conclude $\operatorname {res}^J_M$ is surjective.

Exchanging $M$ for a conjugate if necessary, we may assume that $J_p \leq M$. As $\operatorname {res}^M_{J_p}$ is injective by induction, it follows that the composition $\operatorname {res}^J_{J_p} = \operatorname {res}^M_{J_p} \circ \operatorname {res}^J_M$ is also injective. On the other hand,

\[ \operatorname{inv}_J H^1(J_p, N) \subseteq \operatorname{inv}_{M} H^1(J_p,N) = \operatorname{res}^M_{J_p} H^1(M,N) \subseteq \operatorname{res}^J_{J_p} H^1(J,N) \]

where the equality above follows from the inductive hypothesis, so that $\operatorname {res}^J_{J_p}$ is surjective.

Otherwise, $q=p$, so that $J_p=Q$ is a Sylow $p$-subgroup of $J$. In this case, $H^1(J/Q,\, N^Q)$ is trivial in (3.2) and so $\operatorname {res}^J_{J_p}$ is injective. As $H^1(Q,\,N)^{J/Q} = \operatorname {inv}_J H^1(Q,\,N)$, it remains to show that this map is surjective. For $M$-invariant $\varphi \in Z^1(J_p,\, N)$, define $\tilde \varphi : J \to N$ by $\tilde \varphi (hm) = \varphi (h)$ for $h\in J_p$ and $m\in M$. Then for any $h,\,h' \in J_p$ and $m,\, m' \in M$, we have $\tilde \varphi (hmh'm') = \varphi (h (h')^{m^{-1}}) = \varphi (h) \varphi ((h')^{m^{-1}})^{h^{-1}}$ $= \varphi (h) \varphi (h')^{m^{-1}h^{-1}} = \tilde \varphi (hm) \tilde \varphi (h'm')^{(hm)^{-1}}$ where the third equality follows from $\varphi$ being $M$-invariant. As $J\cong J_p \rtimes M$, we conclude that $\tilde \varphi \in Z^1(J,\, N)$. Clearly, $\operatorname {res}^J_{J_p} \tilde \varphi \sim \varphi$ so that $\operatorname {res}^J_{J_p}$ is surjective.

For each prime $p$, we may apply proposition 3.1 to the component for $p$ in (3.1) and find that $H^1(J,\,N_p)\cong \operatorname {inv}_J H^1(J_p,\, N_p) \cong \operatorname {inv}_J H^1(J_p,\, N)$ for some $J_p \in \operatorname {Syl}_p(J)$. In particular, it follows that:

Proposition 3.2 Given a group $J$ acting on a nilpotent group $N$ via automorphisms so that $N\rtimes J$ is supersoluble, the restriction maps $\operatorname {res}^J_{J_p}$ induce an isomorphism of pointed sets $H^1(J,\, N) \cong \oplus _{p\in \mathcal {D}} \operatorname {inv}_J H^1(J_p,\, N)$ where $\mathcal {D}$ denotes the set of prime divisors of $\lvert{J}\rvert$ and $J_p\in \operatorname {Syl}_p(J)$ for each $p\in \mathcal {D}$.

We are now prepared to provide a proof of lemma 1.4.

Proof of lemma 1.4 In a supersoluble group $G$, suppose $J$ and $J'$ are locally conjugate complements of a normal nilpotent subgroup $N$. As in lemma 1.1, we have for each prime $p$ that some $J_p\in \operatorname {Syl}_p(J)$ and $J_p'\in \operatorname {Syl}_p(J')$ are conjugate by an element of $N$. Let $\varphi '\in Z^1(J,\,N)$ denote the map corresponding to $J'$. As the isomorphism in proposition 3.2 is induced by restriction maps, it takes the identity $1\in H^1(J,\, N)$ to $\oplus _{p\in \mathcal {D}} 1|_{J_p}$. Thus, as $\varphi '|_{J_p}\sim 1|_{J_p}$ for each $p\in \mathcal {D}$, we may apply proposition 3.2 to conclude $\varphi '\sim 1$ so that $J$ and $J'$ are conjugate.

We now use lemma 1.4 to show:

Proposition 3.3 Let $H$ be a subgroup of some supersoluble $G\cong N \rtimes J$ where $N$ is nilpotent. If for each prime $p$, $H$ contains a conjugate of some $S\in \operatorname {Syl}_p(J)$, then $H$ contains a conjugate of $J$ and so splits over $N\cap H$.

Proof. The hypotheses imply that $H$ supplements $N$ in $G$. We induct on the order of $G$. If $N$ is trivial or if $H$ is a $p$-group, the conclusion follows immediately. If multiple primes divide $\vert{N}\rvert$, then for some prime $p$, $HN_p$ must be a strict subgroup of $G$ for $N_p \in \operatorname {Syl}_p(N)$; otherwise $H$ would contain a Sylow subgroup of $G$ for each prime and we would have $H=G$. Let $p$ be such a prime. Induction in $G/N_p$ implies $J^g \leq HN_p$ for some $g\in G$. Switching to a conjugate of $H$ if necessary, we may assume that $g$ is trivial and apply the inductive hypothesis in $HN_p$ to conclude $J^{g'} \leq H$ for some $g'\in G$. We now proceed under the assumption that $N$ is a $q$-subgroup for some prime $q$.

Let $A \leq N$ be a minimal normal subgroup of $G$; as $G$ is supersoluble, it will have prime order $q$. If $A \leq H$, then in $G/A$, induction implies that $J^gA \leq HA = H$ for some $g\in G$ so that $J^g \leq H$.

Otherwise, $A \cap H$ is trivial. Without loss, $J_q \leq H$ for some $J_q \in \operatorname {Syl}_q(J)$. In $G/A$, induction implies that a conjugate of $JA/A$ is contained in $HA/A$. Let $\overline {K}$ denote this conjugate. Switching to a different conjugate if necessary, we may assume that $J_qA/A \leq \overline {K}$. Let $\varphi : h \mapsto hA/A$ denote the isomorphism from $H$ to $HA/A$ and consider $K = \varphi ^{-1}(\overline K)$. It follows that $J_q \leq K$ and $\vert{K}\rvert = \vert{J}\rvert$ so that $K\leq H$ complements $N$ in $G$. As $N$ is a $q$-group, a Sylow $p$-subgroup of $J$ will be conjugate to a Sylow $p$-subgroup of $K$ for primes $p\ne q$. Lemma 1.4 then implies that $J$ and $K\leq H$ are conjugate in $G$.

We now prove theorem 1.5.

Proof of theorem 1.5 Given $J$, $N$, and $\Omega$ as described in the hypotheses of the theorem, let $G= N \rtimes J$ denote the induced semidirect product and consider $G_\alpha$ for some $\alpha \in \Omega$. As $N$ acts transitively, $G=NG_\alpha$. For each prime $p$, the hypotheses of the theorem imply $(J_p)^{n_p} \leq G_\alpha$ for some $J_p \in \operatorname {Syl}_p(J)$ and $n_p\in N$, so that proposition 3.3 implies $G_\alpha$ contains a conjugate of $J$, say $J^g$ for $g\in G$. It follows that $J$ fixes $\omega =g \cdot a$.

This in turn implies:

Corollary 3.4 Let $G$ be a supersoluble split extension over a nilpotent subgroup $N$. If for each prime $p$ there is a Sylow $p$-subgroup $S$ of $G$ such that any two complements of $S\cap N$ in $S$ are conjugate, then any two complements of $N$ in $G$ are conjugate.

Proof. Suppose arbitrary $J$ and $J'$ complement $N$ in $G$. Then $G$ acts on the cosets $\Omega = G/J'$ in such a way that we may apply theorem 1.5 to infer that $J$ fixes $gJ'$ for some $g\in G$. Consequently, $J$ and $J'$ are conjugate, and we may conclude.

4. Concluding remarks

In their paper, Losey and Stonehewer exhibited a soluble group $G\cong N \rtimes J$ with $N$ nilpotent and $J$ supersoluble and a second complement $J'$ to $N$ in $G$ such that $J$ and $J'$ are locally conjugate but not conjugate [7]. Thus, lemma 1.4 cannot be extended to supersoluble complements of a normal nilpotent subgroup in a soluble group.