# Automorphism groups of a group and of its Fitting subgroup

This was a comment to the answer here . It is one of the series of questions about finite groups with automorphism groups of odd order and would reduce the question to nilpotent groups.

Question. Is it true that a finite group $$G$$ has an automorphism of order 2 iff the Fitting subgroup $$F(G)$$ has an automorphism of order $$2$$.

Geoff Robinson proved in a comment here that one of the implications is true if $$G$$ has odd order. He doubts that "iff" is true but no example was given.

• Couldn't you say which of the implications? Here's the quote: " If $G$ has odd order, it is true that every automorphism of order two of $G$ induces a non-trivial (hence order two) automorphism of $F(G)$" – YCor Feb 18 at 22:41

It can happen that the whole group has no automorphism of order 2 although the Fitting subgroup does:

Let

• $$N$$ be a non-abelian 2-generated $$p$$-group for some prime $$p$$, such that $$N/Z(N)$$ has no automorphism of order $$2$$;
• $$L$$ a nontrivial group of order coprime to $$2p$$ with no automorphism of order $$2$$.

I claim that the wreath product $$G=N\wr L= N^L\rtimes L$$ has no automorphism of order $$2$$, although its Fitting subgroup $$N^L$$ obviously does (permute 2 factors).

Indeed, let $$s$$ be an automorphism with $$s^2=1$$. Then $$s$$ preserves the Fitting subgroup. Also it preserves some conjugate of $$L$$: indeed, consider a $$2|L|$$-Hall subgroup of $$G\rtimes\langle s\rangle$$, and let $$L_1$$ be its subgroup of index $$2$$: then $$L_1$$ is a $$|L|$$-Hall subgroup of $$G$$ and hence is conjugate to $$L$$. So henceforth assume that $$s$$ preserves $$L$$.

Since $$L$$ has no automorphism of order $$2$$, $$s$$ acts as the identity on $$L$$. Hence on $$N^L$$, $$s$$ commutes with the $$L$$-action.

Let now $$t$$ be an arbitrary automorphism of $$N$$ commuting with the $$L$$-action. Let $$N_u$$ be the copy of $$N$$ indexed by $$u\in L$$ in $$N^L$$. For $$u\in L$$ let $$M_u$$ be the projection of $$t(N_1)$$ in $$N$$ (through the $$u$$-th projection to $$N_u$$). Then for $$v\in L$$, $$t(N_v)=t(vN_1)=v(tN_1)$$, so the $$u$$-projection of $$t(N_v)$$ is equal to $$M_{v^{-1}u}$$. Since $$N_1$$ and $$N_v$$ commute for $$v\neq 1$$, we deduce that $$M_u$$ and $$M_{v^{-1}u}$$ commute for $$v\neq 1$$.

Also since the $$N_u$$ generate $$N^L$$, the $$M_u$$ generate $$N$$. Now we will use that $$N$$ is 2-generated (so $$N$$ is a $$p$$-group for some odd prime $$p$$ and $$N/[N,N]N^p$$ has order $$p^2$$). If there exists $$u$$ such that $$M_u=N$$, we deduce that $$M_v$$ is central in $$N$$ for all $$v\neq u$$. Otherwise, there exist $$u\neq v$$, $$x\in M_u$$ and $$y\in M_v$$ such that $$\{x,y\}$$ generates $$N$$; since $$x,y$$ commute, it follows that $$N$$ is abelian, contradiction.

Hence, modulo the center, $$\mathrm{Aut}(N^L)$$ permutes the factors (I used the 2-generation condition there). Now if we restrict to automorphisms commuting with the $$L$$-action, this action on $$L$$ is through the centralizer of the action of $$L_{\mathrm{left}}$$, which is $$L_{\mathrm{right}}$$.

Applied to $$L$$, this shows that for every $$u\in L$$, $$s(N_u)$$ equals $$N_u$$ modulo $$Z(N^L)$$. Composed with the projection to $$N_u$$, this yields an endomorphism $$s_u$$ of $$N_u$$, and we have $$s_u^2$$ induces trivial automorphism of $$N_u/Z(N_u)\simeq N/Z(N)$$. Hence $$s_u$$ induces an automorphism of order $$\le 2$$ of $$N/Z(N)$$, and by assumption, this implies that $$s_u$$ is trivial modulo the center. Hence $$s$$ is trivial modulo the center. But this implies that $$s$$ has order $$p^k$$ dividing the exponent of $$Z(N)$$. So $$s$$ is trivial.