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?

Question

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?

Answer 1

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

Answer 2

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.