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I am being more stupid than usual: embarrassingly so. Let $G$ be a group. Let $\phi:G \rightarrow G$ be an automorphism with inverse $\psi$. Is it true (and why is it true) that $\psi(x)=\phi(x^{-1})$.

Right now, I got myself in a gumption trap and have not thought my way out of it.

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The LHS is a homomorphism from G to G but the RHS is an anti-homomorphism, so... – Qiaochu Yuan Nov 21 '10 at 22:16
up vote 7 down vote accepted

This is not true in general. For instance consider the case of the identity $I_{G}:G \rightarrow G $. It is its own inverse, nevertheless it isn't true that $I_{G}(x) = I_{G}(x^{-1})$ because that would imply that $x = x^{-1}$ for every $x \in G$.

Another counterexample along the same lines. If $G$ is an abelian group then you have available the automorphism $\phi : G \rightarrow G$ given by $\phi (x) = x^{-1}$. Then in this case also $\phi$ is its own inverse and the same argument as before applies.

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Thanks! No wonder I could not prove this. – Scott Carter Nov 21 '10 at 22:07
You're welcome. – Adrián Barquero Nov 21 '10 at 22:10

After brushing off some cobwebs, I thought I might see for which groups the identity can hold.

Claim: Let $G$ be a finite group for which there exists $\phi \in \mathrm{Aut}(G)$ for which $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$. Then $G$ is an elementary abelian 2-group.

Proof: If $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$, then $\phi(\phi(x))=\phi^{-1}(\phi^{-1}(x^{-1}))$ and so on. Hence $x=x^{-1}$ and $x^2=\mathrm{id}$ for all $x \in G$, and so $G$ must be an elementary abelian 2-group. $\square$

In fact, a converse of the above is true, since each elementary abelian 2-group admits a the identity automorphism $\phi$, which satisfies the identity $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$. But, as we will see, not all automorphisms of elementary abelian 2-groups satisfy the identity.

Claim: Let $G$ be a finite group for which each $\phi \in \mathrm{Aut}(G)$ satisfies $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$. Then $G \cong \mathbb{Z}_2$ or $G$ is the trivial group.

Proof: By the previous claim, $G$ must be an elementary abelian 2-group. Applying $\phi$ to both sides of the identity $\phi(x)=\phi^{-1}(x^{-1})$, we find that $\phi(\phi(x))=x^{-1}=x$ for all $x \in G$. Hence $\phi^2$ is the identity automorphism for all $\phi \in \mathrm{Aut}(G)$. Therefore, every automorphism of $G$ must not contain a 3-cycle (in fact t-cycle where $t \geq 3$).

The elementary abelian 2-group of order 4 admits an automorphism that contains a 3-cycle. Consequently, we can prove by induction that elementary abelian 2-group of order $2^a$ for all $a \geq 2$ each admit an automorphism that contains a 3-cycle.

The claim is true for the trivial group and $\mathbb{Z}_2$ by inspection. $\square$

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In the proof of the first claim, you might want to add that you using the fact that $\mathrm{Aut}(G)$ is a finite group, so for every $\psi\in\mathrm{Aut}(G)$ there exists $k\gt 0$ such that $\psi^k = \mathrm{id}$. Also, you are not establishing for which groups the identity can hold, you are finding all groups for which the identity will necessarily hold for all automorphisms. – Arturo Magidin Nov 22 '10 at 2:52


But $\phi(x)^{-1}= \phi^{-1}(x)$, not is necesarily true .

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