# Let $A$ be the group of anti-automorphisms, $B=A\cup \operatorname{Aut}(G)$, prove $B\cong A\times \mathbb{Z}/2\mathbb{Z}$

Let $$A=\{\phi\in S(G):\forall_{a,b\in G}: \phi(ab)=\phi(b)\phi(a)\}$$ be the set of anti-automorphisms of a non-abelian group $$G$$, define $$B=A\cup \operatorname{Aut}(G)$$, with $$\operatorname{Aut}(G)$$ being the group of automorphisms of $$G$$.

Prove that $$B\cong \operatorname{Aut}(G)\times$$ $$\mathbb{Z}/2\mathbb{Z}$$

During my proof that $$B$$ is a group, I discovered some properties that make it logical that $$\mathbb{Z}/2\mathbb{Z}$$ is included here. I also thought that $$|A|=|\operatorname{Aut}(G)|$$ since you can take the same bijections, but apply the necessary conditions. What I noted:

1. $$A$$ and $$\operatorname{Aut}(G)$$ are disjoint
2. $$\phi,\psi\in A \implies \phi\circ\psi \in\operatorname{Aut}(G)$$
3. $$\phi\in A, \psi\in\operatorname{Aut}(G)\implies \phi\circ\psi\in A$$ and $$\psi\circ\phi\in A$$

This gives us that if you compose an even number of elements from $$A$$, you will end up with an element from $$\operatorname{Aut}(G)$$, and if you compose an uneven number of elements from $$A$$, you will end up with another element in $$A$$. This gives me an intuitive feeling that $$\mathbb{Z}/2\mathbb{Z}$$ is involved, but I can't quite find the right bijection to completely prove the statement.

Is this the way to go about it, or should is there a different approach?

• Do the anti-automorphisms form a group? Jun 16, 2018 at 12:11
• No, because a product of two elements from $A$ leaves an element in $\operatorname{Aut}(G)$ which is disjoint from $A$. Only if $G$ is abelian, we have that $A$ is a group, in fact, $A=\operatorname{Aut}(G)$. I think I forgot to add the extra restriction that $G$ must not be abelian.
– Marc
Jun 16, 2018 at 12:15
• Then what do you mean by "let $A$ be the group of anti-automorphisms" and $B \cong \color{red}{\mathbf{A}} \times \Bbb Z/2\Bbb Z$? Jun 16, 2018 at 12:16
• @KennyLau you are right. It should be $\operatorname{Aut}(G)$ instead of $A$, which I corrected now.
– Marc
Jun 16, 2018 at 12:17
• No, you have not corrected it. Jun 16, 2018 at 12:19

$\mathrm{Aut}(G)$ is a normal subgroup of $B$. Now let $\eta: G \to G$ denote the inverse map $\eta(g) = g^{-1}$. Then $\eta$ is an anti-automorphism of order dividing 2, and for for any automorphism or anti-automorphism $\varphi$, $\varphi\eta = \eta\varphi$. Further, $A = \eta \mathrm{Aut}(G)$, so $B$ is the internal direct product $\mathrm{Aut}(G) \times \langle \eta \rangle$ if and only if $\eta \notin Aut(G)$. So $B = Aut(G) \times \langle \eta \rangle$ if $G$ is nonabelian, and $B = Aut(G)$ if $G$ is abelian. $G$ nonabelian also implies $\eta \neq 1$, so then $\eta$ has order two.
Edit: took into account if $G$ is abelian