Proof verification: $(xy)^2=xy$ for all $x,y$ in a group implies the group is Abelian

Question

$(xy)^2=xy$ for all $x,y$ in a group implies the group is Abelian

Proof:

$(xy)^2=xy \implies (xy)^{-1}(xy)^2 = (xy)^{-1}(xy) \implies xy = e$ where $e$ is the identity of the group. Since $x,y$ is arbitrary, it is also true that $yx=e$. Hence $yx=xy$. $\square$

Is this proof correct? Thanks

Edit: This question is from chapter 3 problem 52 of Abstract Algebra Theory and Applications by Judson

This is correct, but the statement you’re proving is somewhat strange. You can determine the group much more precisely than just noting that it is abelian. If this is an exercise: Are you reading it correctly? — Eike Schulte, Jan 13 '22 at 13:16
If $xy=e$ for all $x,y$, what can you deduce by putting $y=e$? — TonyK, Jan 13 '22 at 13:37
The usual formulation (for this exercise) is this one. You should clarify this. For $y=e$ it just says $x^2=x$ for all $x$, which is $x=e$ for all $x$. This seems too trivial. — Dietrich Burde, Jan 13 '22 at 15:33
Note this isn't in the 2021 version (the closest is exercise 51). Which version of the text are you using? — Brian Moehring, Jan 14 '22 at 00:36
@BrianMoehring Ah... sorry. I’m using the August 27, 2010 version — Math55, Jan 14 '22 at 10:56

score 2 · Accepted Answer · edited Jan 14 '22 at 00:14

2

Your proof is right, but as Eike Schulte said, it is somehow strange. I strongly suspect your statement was $\forall x\in G, x^2=x$. I'll prove that statement. (In fact, your statement and my statement is equivalent. why?)

If $x,y\in G$, $$\begin{align} xxyy&=x^2y^2\\ &=xy\\ &=(xy)^2\\ &=xyxy, \end{align}$$ then $xy=yx$.

edited Jan 14 '22 at 00:14

Shaun

47,747

answered Jan 13 '22 at 13:31

1

More simply: $x^2=x\implies x^{-1}x^2=x^{-1}x\implies x=e$. So the group has only one element (and is therefore abelian). – TonyK Jan 13 '22 at 13:39

score 2 · Answer 2 · answered Jan 14 '22 at 00:22

Theorem: The only idempotent of a group $G$ is the identity $e$.

Proof: Suppose $a^2=a\in G$. Then $aa=a=ae$, so, multiplying on the left by $a^{-1}$, we get $a=e$. $\square$

In your question, since $e\in G$, we have $y^2=(ey)^2=ey=y$ for all $y\in G$. Thus $y=e$, so $G$ is trivial and hence abelian.

Proof verification: $(xy)^2=xy$ for all $x,y$ in a group implies the group is Abelian

2 Answers2