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This question already has an answer here:

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!

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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.

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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.

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