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Let $(G,•)$ be a group in which $(a•b)^3 = a^3•b^3$. Prove that $H=\{x^3 : x \in G\}$ is a normal subgroup of $G$.

I assumed a homomorphism from $G$ to $G$ by $f(x) = x^3$. Now $\operatorname{Im}f$ is $H$. This is where I got stuck. How can I show that here $\operatorname{Im}f$ is a normal subgroup of $G$? Please help.

Shaun
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Abhisek Mukherjee
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  • (Almost) the same question is answered here. – Dietrich Burde Mar 21 '18 at 15:21
  • $g^{-1}x^3g=(g^{-1}xg)^3$ – Matthew Towers Mar 21 '18 at 15:23
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    @Surb No...the kernel of a homomorphism is always normal. The range of a homomorphism is cerainly not necessarily normal! If this was the case then every subgroup of every group would be normal, as an embedding of $H$ as a subgroup of a group $G$ is simply an injective homomorphism from the abstract group $H$ into $G$. – user1729 Mar 21 '18 at 15:27

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Let $a,b\in H$ so that $a=x^3$ and $b=y^3$. Since $(ab^{-1})=x^{3}y^{-3}=(xy^{-1})^3\in H$, $H$ is a subgroup.

Let $g\in G$. Then $gag^{-1}=gx^3g^{-1}=(gxg^{-1})^3\in H \implies H\lhd G$.

mesel
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