If $x\mapsto x^{-1}$ is a group morphism, then $x\mapsto x^2$ is a group morphism

Question

Prove that, $$f:(G,·)\longrightarrow(G,·) \text{ with }f(x)=x^{-1} \text{, morphism } \implies g:(G,·)\longrightarrow(G,·) \hspace{5mm}\text{ with }g(x)=x^2 \text{, morphism}$$ Where morphism means a morphism between groups.

To prove this statement I have to show that $g(xy)=g(x)g(y)$, i.e. $$xyxy=x^2y^2$$ using the fact that $x^{-1}y^{-1}=y^{-1}x^{-1}$ (since $\varphi^{-1}$ is a morphism).

My attempt, $$xyxy=x^{2}y^{2}\iff f(xyxy)=f(x^2y^2)\iff y^{-1}x^{-1}y^{-1}x^{-1}=y^{-2}x^{-2}\iff x^{-1}y^{-1}=y^{-1}x^{-1}\hspace{2mm}\blacksquare$$ Is my proof correct?

I have to point out that this question has already been answered here, although it didn't work on the implication (ii) $\implies$ (iii) which is the one I'm working on. So please, don't close this one.

Thanks in advance.

Answer 1

Probably the easiest way to see this is to prove that if $x\mapsto x^{-1}$ is an automorphism (note it is clearly its own inverse), then $G$ is abelian. This is easy to see since $\phi((xy)^{-1})=yx$ while $\phi(x^{-1})\phi(y^{-1})=xy$, so we have that $xy=yx$. That is if $x\mapsto x^{-1}$ is an automorphism, then $G$ is abelian. From this it is clear that $\psi(xy)=(xy)^2=x^2y^2=\psi(x)\psi(y)$ is a homomorphism.

Answer 2

Your bidirectional implications are fine - the only thing to plug is to mention that $f$ is a bijection (regardless of whether or not it is a group homomorphism), therefore the first bidirectional implication holds, specifically $f(xyxy) = f(x^2y^2) \implies xyxy = x^2y^2$ requires injectivity of $f$.

So your proof is mostly fine, but a bit convoluted. Note that $g(xy)=g(x)g(y)$ is equivalent to $xyxy=x^2y^2$, which is equivalent to $xy=yx$ by multiplying on the left by $x^{-1}$ and on the right by $y^{-1}$. So $g$ being a homomorphism is equivalent to every pair of elements of $G$ commuting.

The same trick can be used to show that $f$ being a homomorphism is equivalent to $G$ being abelian.

You can turn these into a cycle of implications as in your linked question, but I don't think much is really gained by doing so in lieu of proving both equivalences with $G$ being abelian.