Prove that x and y commute

Question

Suppose G is a group with x and y as elements. Show that $(xy)^2 = x^2 y^2$ if and only if x and y commute.

My very basic thought is that we expand such that $xxyy = xxyy$, then multiply each side by $x^{-1}$ and $y^{-1}$, such that $x^{-1} y^{-1} xxyy = xxyy x^{-1}$ , and therefore $xy=xy$.

I realize that this looks like a disproportionate amount of work for such a simple step, but that is what past instruction has looked like and that is perhaps why I am confused. Moreover, "if and only if" clauses have always been tricky for me since I took Foundations of Math years ago, but if I remember correctly, the goal here should be to basically do the proof from right to left and then left to right, so to speak. Anyhow, I think that I am overthinking this problem.

Answer 1

Hint

$$(xy)^2=x^2y^2\iff xyxy=xxyy\iff x^{-1}xyxy=x^{-1}xxyy\iff yxy=xyy.$$

Can you finish?

Answer 2

If $x$ and $y$ commute, clearly we have $$ xyxy=xxyy=x^2y^2 $$ if instead $$ xyxy=x^2y^2 $$ then hitting the left side with $x^{-1}$ and the right with $y^{-1}$ yields $$ xy=yx $$

Answer 3

The proof can be written fairly concisely as follows:

\begin{align} (xy)^2 = x^2 y^2 &\iff (xy)(xy) = (xx)(yy) && (\text{expand both sides}) \\ &\iff xyxy = xxyy && (\text{association}) \\ &\iff x^{-1} (xyxy) y^{-1} = x^{-1} (xxyy) y^{-1} &&(\text{cancelation}) \\ &\iff (x^{-1} x) yx (y y^{-1}) = (x^{-1} x)xy(y y^{-1}) &&(\text{association}) \\ &\iff yx = xy. &&(\text{def'n of inverses}) \end{align}

Answer 4

You're right that you should do the proof from right to left and then left to right. But make sure you know what “left” and “right” are.

“left” is the equation $(xy)^2=x^2y^2$, while “right” is the statement “$x$ and $y$ commute.” You can write the last statement as the equation $xy=yx$.

This means you want to show:

1. $(xy)^2 = x^2y^2 \implies xy =yx$
2. $xy=yx \implies (xy)^2 = x^2y^2$

For the first, you can expand so that (better word choice than such that, in my opinion) $xyxy = xxyy$. Multiply on the left by $x^{-1}$ and on the right by $y^{-1}$, so that $yx=xy$.

For the second, $xy=yx$ means that you can swap $x$ and $y$ in $xxyy$. So $x^2y^2=xyxy=(xy)^2$.