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Let $G$ be a group where $(ab)^3=a^3b^3$ . Prove that $\{ x^3 \mid x \in G \}$ is a normal subgroup of $G$.

I just don't know how to start. I proved it is a subgroup (was pretty easy) but how to prove it's normal? Thanks in advance for your help!

Shaun
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Matteo Rossi
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  • You need to prove that every conjugate of a cube is a cube - is that true or not? If it is true, then the collection of cubes is closed under conjugation ... – Mark Bennet Feb 20 '20 at 18:31
  • Let $H={x^3|x\in G}$. Note that $e\in G$ is in $H$ and if $a,b\in H$, then there are $x,y\in G$ such that $a=x^3$ and $b=y^3$. Then $ab=x^3y^3=(xy)^3\in H$. Therefore, $H$ is a subgroup. Now, for $a\in G$ and $x^3\in H$, we have that $ax^3a^{-1}=(axa^{-1})(axa^{-1})(axa^{-1})=(axa^{-1})^3\in H$. –  Feb 20 '20 at 18:31
  • Two ideas relevant to the proof and useful to remember were: The characterization of subgroups as containing the identity and closed under the operation, and that conjugation commutes with powers $ax^na^{-1}=(axa^{-1})^n$ due to that trick of inserting $aa^{-1}$ in between. –  Feb 20 '20 at 18:34
  • @infinity Why would you delete your answer? –  Feb 20 '20 at 18:36
  • Actually, here, the fact that it is normal is the trivial part: this set is always invariant under conjugation. It is the fact that it is a subgroup that is novel. For most groups, the set of cubes is not a subgroup. – Arturo Magidin Feb 21 '20 at 04:34

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Denote $H = \{x^3:x\in G\}$.

You showed already that this is a subgroup of $G$.

Now we want to show that $gHg^{-1} = H$ for all $g\in G$.

So let $g\in G$. if $x\in gHg^{-1}$ then $x= gy^3g^{-1} $ for some $y\in G$.

But then $(gyg^{-1})^3 =x$ so $x\in H$.

So, you don't need the requirement $(ab)^3 = a^3 b^3$ for the normality part.

infinity
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