# Let $G$ be a group in which $a^2=e$ for all elements of $a$ of $G$. Show that $G$ is Abelian. [duplicate]

Let $G$ be a group in which $a^2=e$ for all elements of $a$ of $G$.

Show that $G$ is Abelian.

I need help on this problem. Appreciated!

## marked as duplicate by Myself, N. S., Cameron Buie, rschwieb, Martin BrandenburgApr 26 '13 at 18:02

• – MJD Apr 26 '13 at 17:52
• Can we get this question and solution on the front page of a national newspaper (or website) somewhere? I am so tired of hearing this question... :P Or maybe a bot that is specifically looking for this question to appear. – rschwieb Apr 26 '13 at 17:52
• Actually it also holds for monoids. ;) – Martin Brandenburg Apr 26 '13 at 18:02

Consider two elements $g,h \in G$. You know that $(gh)^2 = e$ and so you get

$$gh gh = e.$$

Applying $h$ to both sides on the right gives you

$$gh g \underbrace{h h}_\text{h^2 = e} = e \cdot h = h.$$

So now you have

$$ghg = h.$$

Aimilarly, applying $g$ to both sides on the right gives you that

$$gh\underbrace{gg}_\text{g^2} = hg,$$

and so you end up with

$$gh = hg,$$

proving $G$ is abelian.

Hint: You want to show that $ab=ba$ for all $a$ and $b$.

It is natural to start from $(ab)(ab)=e$. Use associativity to rewrite this as $a((ba)b)=e$. Now exploit the fact that $a^2=e$, and we are close to the end.