Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

3 Answers 3

up vote 6 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.

share|improve this answer
    
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. $\square$

share|improve this answer
    
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

$\phi(x^{-1})=\phi(x)^{-1}$

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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