# 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.

• I'm not sure if it helps, but one property of the commutator subgroup (which is trivial if and only if your group is abelian) is that for a homomorphism $f$ and two elements $g,h$ of your group, you have $f([g,h]) = [f(g), f(h)]$ where $[a,b] = aba ^{-1}b^{-1}$ is the commutator – user2566092 Mar 7 '15 at 21:07
• Is $G$ a finite group? – punctured dusk Mar 7 '15 at 21:17
• I am pretty sure this problem also appears in Herstein's classic "Topics in Algebra". – Prism Mar 8 '15 at 1:36

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.

• Is $G$ finite in your proof? – user149418 Mar 7 '15 at 22:17
• @user149418 Yeah, I'm supposing the hypothesis $3\nmid |G|$ is implicitly saying $G$ is finite. – Ben West Mar 7 '15 at 22:19
• I think it is, otherwise $\phi$ might not be an isomorphism. – krirkrirk Mar 7 '15 at 22:22
• Could you please explain how did you conclude that $\phi$ is an automorphism from the fact that 3 does no divide $|G|$? – user557550 Mar 20 at 0:46
• @MariyaKav If $g\in\ker\phi$, $\phi(g)=g^3=1$, so $g$ has order dividing $3$. But $3\nmid|G|$, so no element of $G$ has order $3$ by Lagrange. Necessarily such $g$ has order $1$, so $g=1$. Thus $\phi$ is injective as it has trivial kernel, hence is bijective as an endomorphism of a finite group, hence an automorphism. – Ben West Mar 20 at 5:59

$\phi$ 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.