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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?

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  • $\begingroup$ Do the anti-automorphisms form a group? $\endgroup$
    – Kenny Lau
    Jun 16, 2018 at 12:11
  • $\begingroup$ 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. $\endgroup$
    – Marc
    Jun 16, 2018 at 12:15
  • $\begingroup$ Then what do you mean by "let $A$ be the group of anti-automorphisms" and $B \cong \color{red}{\mathbf{A}} \times \Bbb Z/2\Bbb Z$? $\endgroup$
    – Kenny Lau
    Jun 16, 2018 at 12:16
  • $\begingroup$ @KennyLau you are right. It should be $\operatorname{Aut}(G)$ instead of $A$, which I corrected now. $\endgroup$
    – Marc
    Jun 16, 2018 at 12:17
  • $\begingroup$ No, you have not corrected it. $\endgroup$
    – Kenny Lau
    Jun 16, 2018 at 12:19

1 Answer 1

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$\mathrm{Aut}(G)$ 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

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