If G is a group and $a^2=e$, then G is abelian.

Question

My Proof:

$$x=x$$ $$ex=xe$$ $$y^2x=xy^2$$ $$y^{-1}y^2x=xy^2y^{-1}$$ $$yx=xy$$

Is this right? Thanks!

In the line before the last, you multiplied by $y^{-1}$ on the left on one side and on the right on the other side. — lulu, Jan 17 '24 at 20:09
$y^2x=xy^2\Rightarrow y^{-1}y^2x=xy^2y^{-1}$ is a mistake of algebra. — Sassatelli Giulio, Jan 17 '24 at 20:11
This is not a "check-my-proof" site. You should follow the instructions in the solution-verification tag (which was proper for this post and I added): ""Is my proof correct?" is off topic (too broad, missing context). Instead, the question must identify precisely which step in the proof is in doubt, and why so. " Your post is either inappropriate, or a duplicate. — Arturo Magidin, Jan 17 '24 at 20:11
If that's what you mean, Arthr, then that's what you should write. But it's a moot point, since apparently the question's a duplicate. — Gerry Myerson, Jan 17 '24 at 22:11

score 0 · Answer 1 · answered Jan 17 '24 at 20:19

0

$x^2=1$ imply $x=x^{-1}$

So $xy=x^{-1}y^{-1}=(yx)^{-1}=yx$

answered Jan 17 '24 at 20:19

F13

4

This question seems not to meet the standards for the site. Instead of answering it, it would be better to look for a good duplicate target, or help the user by posting comments suggesting improvements. Please also read the meta announcement regarding quality standards. – Shaun Jan 17 '24 at 20:28

1 Answers1