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I want to show that $\forall x$,$y\in G$, $(xy)^2=x^2y^2 \iff G $ is an abelian group.

Proof of the backward direction seems easy enough:

Assume $G$ is an abelian group. Then consider $(xy)^2=(xy)(xy)$. By associativity, $(xy)(xy)=xyxy$. Since $G$ is abelian, $xyxy=xxyy=x^2y^2$.

Proof of the forward direction I attempt by contrapositive:

That is, assume $G$ is not an abelian group. Then, consider $(xy)^2=xyxy$ as above. Yet, $xyxy\neq xxyy=x^2y^2$ since we no longer have commutativity.

Is this proof by contrapositive completely correct?

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    $\begingroup$ No, it isn't. The inequality you state may be not true for all $x,y \in G$. $\endgroup$ Oct 4 '16 at 22:48
  • $\begingroup$ No. You have to show commutativity like $ab =ba$. $\endgroup$ Oct 4 '16 at 22:49
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    $\begingroup$ It's simpler than you are making it. if $g=h$ then $x^{-1}gy^{-1}=x^{-1}hy^{-1}$. Now apply this to $(xy)^2=x^2y^2$ $\endgroup$
    – lulu
    Oct 4 '16 at 22:50
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    $\begingroup$ Saying that $G$ is not abelian means that $xy\ne yx$ for some $x$ and $y$, not that $xy\ne yx$ for all $x$ and $y$. For instance, if $y=1$, then $xy=yx$. $\endgroup$
    – egreg
    Oct 4 '16 at 22:52
  • $\begingroup$ In other words you are assuming that the centre of the group is trivial.Which may not be the case every time. $\endgroup$ Oct 4 '16 at 23:50
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suppose $xyxy = xxyy$

$x^{-1}(xyxy)y^{-1} = x^{-1}(xxyy)y^{-1}$

$yx = xy$

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