If $\phi(g)=g^3$ is a homomorphism and $3 \nmid |G|$, $G$ is abelian. As the title suggests. Let $G$ be a group, and suppose the function $\phi: G \to G$ with $\phi(g)=g^3$ for $g \in G$ is a homomorphism. Show that if $3 \nmid |G|$, $G$ must be abelian.
By considering $\ker(\phi)$ and Lagrange's Theorem, we have $\phi$ must be an isomorphism (right?), but I'm not really sure where to go after that.
This is a problem from Alperin and Bell, and it is not for homework.
 A: Note that $(gh)^3=\varphi(gh)=\varphi(g)\varphi(h)=g^3h^3$. This implies $ghghgh=ggghhh$, and hence after cancelling, $hghg=gghh$, or $(hg)^2=g^2h^2$. 
I claim that every element commutes with every square in $G$. Let $x\in G$ be arbitrary, and let $a^2\in G$ be an arbitrary square. Since $\varphi$ is an automorphism (here is where we used the fact that $3\nmid |G|$), $x=\varphi(y)=y^3$ for some $y$. Then
$$
ay^3a^{-1}=(aya^{-1})^3=\varphi(aya^{-1})=a^3y^3a^{-3}
$$
which implies $y^3=a^2y^3a^{-2}$, or $y^3a^2=a^2y^3$. That is, $xa^2=a^2x$, and completes the claim. 
So in particular, $g^2h^2=h^2g^2$. So we get $hghg=(hg)^2=g^2h^2=h^2g^2=hhgg$. Cancelling yields $gh=hg$, so $G$ is abelian. 
A: $\phi$ is an isomoprhism, because $g^{3} = 1 \iff g=1$ for the order of $g\in G$ must divide the order of $G$. 
I guess you have to consider $\phi (gh)$ and $\phi(hg)$. 
If $\forall g, h\in G, \phi(gh)=\phi(hg)$, then $ghg^{-1}h^{-1}\in ker\phi$ thus $gh=hg$ and it's won. But not really sure how to do it. 
