The only fixed-point free automorphism of order $2$ is $\phi(a)=a^{-1}$(in a finite group)

Question

I got the problem in Dummit and Foote's Algebra book to prove if $G$ is a finite group that has an automorphism $\phi$ in which if $a=\phi(a)$ then $a=1$. And which satisfies $\phi(\phi(a))=a$ for all $a$ then $G$ is abelian.

Here is what I did: I first prove the map $\omega(a)=a^{-1}\phi(a)$ is injective(and since the group is finite and the domain is the same as the co-domain this proves it is also bijective):

$a^{-1}\phi(a)=b^{-1}\phi(b)\implies ba^{-1}\phi(a)=\phi(b)\implies ba^{-1}=\phi(b)\phi(a)^{-1}=\phi(ba^{-1})\implies ba^{-1}=1\implies b=a$.

So every element in $G$ is of the form $a^{-1}\phi(a)$. Notice $\phi(a^{-1}\phi(a))=\phi(a^{-1})a$. Which is the inverse of $a^{-1}\phi(a)$ which tells us $\phi(g)=g^{-1}$.

From here we get $\phi(ab)=b^{-1}a^{-1}=a^{-1}b^{-1}=\phi(a) \phi(b)$ multiplying by $a$ and $b$ on both sides gives $ab=ba$ as desired.

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My question is: is what I did correct (especially everything up to the point where I conclude $\phi(a)=a^{-1}$)? Normally I wouldn't ask this, but the fact that it asked me to prove something much weaker instead of characterizing the automorphism uniquely makes me doubt it is OK.

Answer 1

The proof looks good, just for completion I will add a proof that $f: a \rightarrow a^{-1}$ is an isomorfism if and only if $G$ is abelian.

If the group is abelian then $f(ab) = (ab)^{-1} = (ba)^{-1} = a^{-1}b^{-1} = f(a)f(b)$.

If the group is not abelian take $x$ and $y$ that do not commute and let $a=x^{-1},b=y^{-1}$ and notice $f(ab) = (x^{-1}y^{-1})^{-1} = yx \neq xy = f(a)f(b) $