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Suppose that $g^2=e$ for all elements $g$ of a group $G$. Prove that $G$ is commutative.

How would I go about doing this proof?

I understand what it means by $g^2=e$, and a group.

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marked as duplicate by Brett Frankel, Michael Albanese, N. S., Erick Wong, Jacob Black Feb 26 '13 at 1:59

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.

how is this question a duplicate? Does Abelian and commutative mean the same thing? – Username Unknown Feb 26 '13 at 2:00
Yes. A group is abelian if the binary operation is commutative. – Michael Albanese Feb 26 '13 at 2:05
Oh that makes sense. I thought about it but my professor was dodgy about telling me that – Username Unknown Feb 26 '13 at 2:08
up vote 3 down vote accepted

$g^2 = 1$ for all $g \in G \implies g^{-1} = g$.

let $a,b \in G$. We have $ab = (ab)^{-1} = b^{-1} a^{-1} = ba.$ Thus $G$ is abelian.

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Hint: You basically only have one move to make: You know that for any $g, h \in G$ you have $(gh)^2 = e$. Trying expanding playing with that equation.

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Remember what you want to prove is $\forall g, h \in G, gh=hg$, and what you know is $e=hggh = hghg = e$.

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