This is inspired by this question. Is there a description of finite groups without automorphisms of order $2$?

3$\begingroup$ @IgorRivin $C_p\times C_p$? $\endgroup$– Jeremy RickardFeb 1, 2020 at 19:36

3$\begingroup$ @IgorRivin Isn't $C_p \times C_p$ a $p$group with a clear automorphism of order $2$? (Side note: $C_p$ itself has an automorphism of order $2$, too) $\endgroup$– user44191Feb 1, 2020 at 19:55

3$\begingroup$ I guess we do not have a complete classification, but see for example here and here for some previous discussion. $\endgroup$– Mikko KorhonenFeb 1, 2020 at 20:24

1$\begingroup$ Every Abelian group that is not an $\mathbb F_2$ vector space admits inversion as an order2 automorphism. $\endgroup$– LSpiceFeb 1, 2020 at 20:47

6$\begingroup$ There is also a result which states that for "almost all" $p$groups (in some sense of almost all), the automorphism group is a $p$group. So for $p > 2$ almost all $p$groups have automorphism group of odd order. $\endgroup$– spinFeb 1, 2020 at 22:12
1 Answer
New version (existence hinted in previous version): If $G$ is a nontrivial finite (solvable) group of odd order with $\Phi(G) = 1$, then $G$ has an automorphism of order $2$.
It is wellknown and easy to check that $F = F(G)$ is a product of minimal normal subgroups of $G$, each an elementary Abelian $p_{i}$group for some prime $p_{i}$. Also, $F$ is wellknown to be complemented in $G$ in this case (I give a proof for completeness:
Choose a proper subgroup $H$ of $G$ minimal subject to $G = FH$ (such exists because $1 \neq F \not \leq \Phi(G)$). Then $(H \cap F) \leq \Phi(H)$ by minimality of $H$. Also $F \cap H$ is normal in $\langle H,F \rangle = G$, since $F$ is Abelian and $F \lhd G$. If $F \cap H \neq 1$, then there is a maximal subgroup $M$ of $G$ with $G = (F \cap H)M$ since $\Phi(G) = 1$. Then $H = (F \cap H)(M \cap H)$ by Dedekind's modular law. But then $H = H \cap M \leq M$ since $F \cap H \leq \Phi(H)$. But then $G = (F \cap H)M \leq M$, contrary to the fact that $M$ is maximal).
Now $G = FH$ for some subgroup $H$ of ${\rm Aut}(F)$, and the product is semidirect. Thus $G$ is isomorphic to a subgroup of the holomorph $ X = F{\rm Aut}(F)$ (the semidirect product of $F$ with its automorphism group). Here, we have $G \cong F{\rm Aut}_{G}(F)$, where ${\rm Aut}_{G}(F)$ is the subgroup of ${\rm Aut}(F)$ induced by the conjugation action of $G$ on $F$.
Now let $t$ be the central element of ${\rm Aut}(F)$ of order $2$ which inverts $F$ elementwise (note that $t$ is indeed central in ${\rm Aut}(F)$, because $\alpha(f)^{1} = \alpha(f^{1})$ for every $\alpha \in {\rm Aut}(F)$). Then $F\langle t \rangle$ normalizes every subgroup of $X$ containing $F$, so normalizes $F{\rm Aut}_{G}(F) \cong G$.
Now $(F{\rm Aut}_{G}(F))(F \langle t\rangle) = 2F{\rm Aut}_{G}(F)$, so that $t$ induces an automorphism of order $2$ of $F{\rm Aut}_{G}(F) \cong G$ (recall that $t$ already inverts $F$ elementwise). Note that $F{\rm Aut}_{G}(F)$ is of index $2$ in $(F{\rm Aut}_{G}(F))(F \langle t\rangle)$, so is normal in the latter group.


$\begingroup$ @spin: Note that $C_{M}(V) \lhd \langle V,M \rangle = G$. If $C_{M}(V) \neq 1,$ then $C_{M}(V)$ contains a minimal normal subgroup $U$ of $G$. This is different from $V$, since $M \cap V = 1$. This contradicts the fact that $G$ has a unique minimal normal subgroup. $\endgroup$ Feb 2, 2020 at 22:01

$\begingroup$ Of course, thanks. It seems to me this would work for any $G$ odd and $G = V \rtimes M$, $V$ elementary abelian, $V$ faithful $M$module. $\endgroup$– spinFeb 2, 2020 at 23:23

1$\begingroup$ It is also true that if $G$ is not solvable then $Aut(G)$ has an involution. Indeed, $G$ is even, hence $G$ has an involution $x$. Hence we can assume that $x$ is central. The group $G/Z(G)$ is not solvable, so it also contains an involution. Hence there exists $y\in G\setminus Z(G)$ such that $y^2\in Z(G)$. Then the inner automorphism of $y$ is an involution in $Aut(G)$. $\endgroup$– user6976Feb 3, 2020 at 18:51

1$\begingroup$ I don't think the iff works, though I can't think of an example where it fails at the moment. If $G$ has odd order, it is true that every automorphism of order two of $G$ induces a nontrivial (hence order two) automorphism of $F(G)$. Hence it is true that if ${\rm Aut}(F(G))$ has odd order, then ${\rm Aut}(G)$ has odd order, when $G$ itself has odd order. $\endgroup$ Feb 4, 2020 at 16:58