2

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.

amWhy
  • 209,954
combo student
  • 1,271
  • 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 02 '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. – amWhy Oct 02 '16 at 16:45
  • Being closed as a duplicate (in this case, the duplicate being http://math.stackexchange.com/questions/238171/prove-that-if-g2-e-for-all-g-in-g-then-g-is-abelian, combo student, does not reflect badly on you or your question. You did a nice job with your question, showing effort, and being articulate. – amWhy Oct 02 '16 at 16:48

2 Answers2

2

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 $(xy)^2=e\implies xyxy=e\implies yxy=x\implies xy=yx$.

Teoc
  • 8,700
  • 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 02 '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 02 '16 at 17:16
1

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 $$xey=xy=x(xyxy)y=x^2yxy^2=eyxe=yx.$$

Bernard
  • 175,478