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Let G be a finite group whose order is not divisible by 3. Suppose that $(ab)^3 = a^3b^3$ for all a, b in G. Prove G must be abelian.

In the proof here, there comes a step, where we have established that $a^2b^3 = b^3a^2$. And previously, we have also established that "Every element of G can be uniquely represented as a cube".

Thus we conclude that $a^2b^2 = b^2a^2$

This is the part I don't understand. If $a^2b^3 = b^3a^2$ and $a^2b = ba^2$, then how does this implies to $a^2b^2 = b^2a^2$

Shaun
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latus_rectum
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  • If every element is a cube then $b^2=c^3$ for some $c$. So then $a^2c^3=c^3a^2\implies a^2b^2=b^2a^2$. – lulu Sep 13 '20 at 13:27

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Because $a^2b=ba^2$ gives $$b=a^{-2}ba^2,$$ which gives $$b^2=a^{-2}b^2a^2$$ and from here $$a^2b^2=b^2a^2.$$

Michael Rozenberg
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