# Let $G$ be a group. Prove that IF $x^2 = e$ for all $x \in G$, then $G$ is abelian. [duplicate]

Let $G$ be a group. Prove that IF $x^2 = e$ for all $x \in G$, then $G$ is abelian.

My attempt:

$x^2 = e$

$x = x^{-1}$

$xx^{-1}=x^{-1}x^{-1}$

$e = x^{-1}x^{-1}$

$e = (xx)^{-1}$

$e = (x^2)^{-1}$

$e^{-1} = ((x^2)^{-1})^{-1})$

$e = x^2$

Hence, $G$ is abelian.

## marked as duplicate by Stefan4024, Namaste abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 2 '16 at 16:39

• For showing $G$ is Abelian you have to show that $aba^{-1}b^{-1}=e$ for all $a,b \in G$ but your proof don't really prove anything. – Arpit Kansal Oct 2 '16 at 16:24
• The first question marked as a "duplicate", @Stefan, is not a duplicate of this question. Did you actually go and read that post? Or did you mark as a duplicate based on similarities (both about groups of order $2$)? Please actually go to/click on any question you think might be a dupe, and read the post, *before" closing as a dupe. – Namaste Oct 2 '16 at 16:45
• Being closed as a duplicate (in this case, the duplicate being math.stackexchange.com/questions/238171/…, combo student, does not reflect badly on you or your question. You did a nice job with your question, showing effort, and being articulate. – Namaste Oct 2 '16 at 16:48

Your method only proves $e=x^2\implies x^2=e$. What you want to prove instead is $xy=yx$ for all x, y in G.
You can prove that it is abelian by $(xy)^2=e\implies xyxy=e\implies yxy=x\implies xy=yx$.
• so then would something like... $a= a^{-1}, b= b^{-1}$ $ab =a^{-1}b^{-1}= (ab)^{-1} = (ba)^{-1}= b^{-1}a^{-1}$ work? – combo student Oct 2 '16 at 16:28
• @combo student The inverse law is $a^{-1}b^{-1}=(ba)^{-1}$. In your method, you use the inverse law wrong, then you assume abelianess in your second to last step. Instead you could have (ba)^{-1}=ba by x^2=e. – Teoc Oct 2 '16 at 17:16
I think st sentence is $\;$‘Let $G$ be a group’.
Start from (xy)^2=xyxy=e, and multiply both sides by $x$ on the left, by $y$ on the right. You get $$xey=xy=x(xyxy)y=x^2yxy^2=eyxe=yx.$$