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This question already has an answer here:

Let $G$ be a group. Show that if every non-identity element in $G$ has order $2$ then $G$ is abelian.

Proof:

Let $a,b $ be non-identity elements in $G$. Since $|a|=|b|=2$ , that means $ab=babaab$ $=$ $ba$.

Is the proof correct? How can I improve it?

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marked as duplicate by rschwieb abstract-algebra May 6 at 13:46

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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    $\begingroup$ The proof is correct as written. $\endgroup$ – rubikscube09 May 4 at 21:04
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    $\begingroup$ I must be tired by I don't see how to justfify the first equality. $\endgroup$ – elidiot May 4 at 21:08
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    $\begingroup$ @elidiot (baba)ab=(e)ab=ab. bab(aa)b=babb=ba $\endgroup$ – topologicalmagician May 4 at 21:09
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    $\begingroup$ It uses $|ba|=2$ on the left. $\endgroup$ – Berci May 4 at 21:10
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    $\begingroup$ @rschwieb thanks, i'll keep that in mind next time. $\endgroup$ – topologicalmagician May 7 at 14:07
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Alternative proof : $a^2=1$ so every element is its one inverse.

So, $(ab)^{-1}=ab$, but $(ab)^{-1}=b^{-1}a^{-1}=ba$ by using twice again the remark.

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Your proof is correct.

You can improve it by making the derivations clearer (as you have done in the comments).

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