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Prove that if $G$ is a group with $(ab)^{2}=a^{2}b^{2}$ for all $a,b\in G$, then $G$ is abelian.

I am not sure how to prove this. This is my work so far:

Proof so far: $G$ is abelian iff $ab=ba$. If $(ab)^{2}=a^{2}b^{2}$ for all $a,b\in G$, then $(ae)^{2}=a^{2}e^{2}=a^{2}$ or $a=a$ for all $a\in G$.

I am not sure whether I am approaching this correctly at this point, since it looks a little silly to me. If this is the right way, I am not sure how to continue.

  • $a = a$ is something you already knew about any group. However, you should use the special relation you've been given more generally. For instance, what can you massage $(ab)^2 = a^2b^2$ into? – Arthur Jan 29 '18 at 15:35
  • Proving $a=a$ for all $a \in G$ is not new information. – quasi Jan 29 '18 at 15:35
  • I don't believe that your approach leads to anywhere. Perhaps you should try to manipulate the asserted identity $(ab)^2=a^2b^2$ and try to end up with $ab=ba$. Keep in mind that every element has an inverse in a group. – asdq Jan 29 '18 at 15:35
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    If $abab=aabb$,cancel $a$ on the left and $b$ on the right... –  Jan 29 '18 at 15:35

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