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?

  • $\begingroup$ Do the anti-automorphisms form a group? $\endgroup$ – Kenny Lau Jun 16 '18 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 '18 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 '18 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 '18 at 12:17
  • $\begingroup$ No, you have not corrected it. $\endgroup$ – Kenny Lau Jun 16 '18 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.