Given that $f:G\to G$ on a group $G$, defined by $f(x)=x^3$, is an isomorphism, how do I show that $G$ is abelian? Given that $f:G\to G$ on a group $G$, defined by $f(x)=x^3$, is an isomorphism, how do I show that $G$ is abelian, please?
 A: Note that
$$
\forall a,b \in G: \quad ababab = (ab)^{3} = a^{3} b^{3} = aaabbb.
$$
Hence,
$$
\forall a,b \in G: \quad baba = aabb, \quad \text{or equivalently}, \quad (ba)^{2} = a^{2} b^{2}.
$$
Using this fact, we obtain
\begin{align}
\forall a,b \in G: \quad (ab)^{4} &= [(ab)^{2}]^{2} \\
                                  &= [b^{2} a^{2}]^{2} \\
                                  &= (a^{2})^{2} (b^{2})^{2} \\
                                  &= a^{4} b^{4} \\
                                  &= aaaabbbb.
\end{align}


*

*On the other hand,
\begin{align}
\forall a,b \in G: \quad (ab)^{4} &= abababab \\
                                  &= a (ba)^{3} b \\
                                  &= a b^{3} a^{3} b \\
                                  &= abbbaaab.
\end{align}

*Hence, for all $ a,b \in G $, we have $ aaaabbbb = abbbaaab $, which yields
$$
f(ab) = a^{3} b^{3} = b^{3} a^{3} = f(ba).
$$
As $ f $ is injective, we conclude that $ ab = ba $ for all $ a,b \in G $.Hence $G$ is an abelian group
A: Answer of the changed question: Note that for any $a,b \in G$, $xy=f(x^{-1})f(y^{-1})=f(x^{-1}y^{-1})=f((yx)^{-1})=yx$.Hence $G$ is abelian.
