# Automorphism groups of odd order

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

• @IgorRivin $C_p\times C_p$? Feb 1, 2020 at 19:36
• @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) Feb 1, 2020 at 19:55
• I guess we do not have a complete classification, but see for example here and here for some previous discussion. Feb 1, 2020 at 20:24
• Every Abelian group that is not an $\mathbb F_2$ vector space admits inversion as an order-2 automorphism. Feb 1, 2020 at 20:47
• 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.
– spin
Feb 1, 2020 at 22:12

New version (existence hinted in previous version): If $$G$$ is a non-trivial finite (solvable) group of odd order with $$\Phi(G) = 1$$, then $$G$$ has an automorphism of order $$2$$.

It is well-known 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 well-known 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 $$|(F{\rm Aut}_{G}(F))(F \langle t\rangle)| = 2|F{\rm Aut}_{G}(F)|$$, 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.

• How do we see that $C_M(V) = 1$?
– spin
Feb 2, 2020 at 21:13
• @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. Feb 2, 2020 at 22:01
• 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.
– spin
Feb 2, 2020 at 23:23
• 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 $Aut(G)$.
– user6976
Feb 3, 2020 at 18:51
• 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 non-trivial (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. Feb 4, 2020 at 16:58