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

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.

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.

• The proof looks fine to me. The final step (if $a\mapsto a^{-1}$ is a homomorphism then $G$ is abelian) may have appeared in an earlier exercise (without the assumption oif finiteness) – Hagen von Eitzen Jul 6 '14 at 16:25
• Yes, I'm pretty confident on the last part, it was the earlier part I wasn't completely sure on. Thank you Hagen. – Yorch Jul 6 '14 at 16:30
• @HagenvonEitzen could you promote comment to answer? – Yorch Jan 12 '15 at 5:42