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.

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    $\begingroup$ 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 $F(G)$" $\endgroup$ – YCor Feb 18 at 22:41

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


  • $N$ 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.

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