Mathematics Asked by latus_rectum on December 19, 2020
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 $a^2b^2 = b^2a^2$
Because $a^2b=ba^2$ gives $$b=a^{-2}ba^2,$$ which gives $$b^2=a^{-2}b^2a^2$$ and from here $$a^2b^2=b^2a^2.$$
Correct answer by Michael Rozenberg on December 19, 2020
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