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Prove that if in a group $(ab)^2= a^2 b^2$ then the group is commutative.
I am having a hard time doing this. Here is what I have so far:
Proof:
$a^2 b^2= a^1 a^1 b^1 b^1$
=$aa^{-1}bb$
=ebb
Hence,$aa^{-1}=e$
I am stuck, I do not know if this is the right process in proving this

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    Is $(ab)^2=a^2b^2$ for all $a,b$ elements of the group? Or is the group generated by $a,b$ with the relation $(ab)^2=a^2b^2$? – 2017-01-21
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    it is generated by a,b with relation to $(ab)^2=a^2 b^2$ – 2017-01-21
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    if only I had a penny for every time this question has been asked. – 2017-01-21

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$aabb=(ab)^2=abab$ implies that $a^{-1}aabbb^{-1} = a^{-1}ababb^{-1}$. So, $eabe = ebae$. Hence $ab=ba$.

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    so I was on the right track? I just got to work with the $(ab)^2$ @Dan Rust – 2017-01-21
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    Well you didn't use the left hand side of the relation that you were given, so you obviously needed to use that somewhere. – 2017-01-21
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    thanks. Im still learning how to properly do proofs. – 2017-01-21
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    Good luck. With a little bit of practice, you see chains of reasoning like above very quickly and naturally. – 2017-01-21
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    one more thing @Dan Rust, the answer you gave me is commutative right? because commutative is usually presented as a + sign – 2017-01-21
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    A group $(G,\bullet)$ is commutative/abelian if for all $a,b\in G$ $a\bullet b = b \bullet a$. Obviously we can ommit the $\bullet$ notation altogether giving $ab=ba$ or, as you suggest, use additive notation with $+$ giving $a+b=b+a$. It's just notation though. We could talk about the integers with addition and write $nm = n+m$ if we really wanted, as long as the reader understood what we meant. – 2017-01-21