Proving G is commutative. [duplicate]

Question

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• Prove that if $g^2=e$ for all g in G then G is Abelian. 3 answers

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.

Answer 1

$g^2 = 1$ 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.

Answer 2

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.

Answer 3

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