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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how is this question a duplicate? Does Abelian and commutative mean the same thing? – Username Unknown Feb 26 '13 at 02:00
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Yes. A group is abelian if the binary operation is commutative. – Michael Albanese Feb 26 '13 at 02:05
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Oh that makes sense. I thought about it but my professor was dodgy about telling me that – Username Unknown Feb 26 '13 at 02:08
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$g^2 = 1$ 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.
jim
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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.
Jim
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Remember what you want to prove is $\forall g, h \in G, gh=hg$, and what you know is $e=hggh = hghg = e$.
NECing
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