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All groups of order 6 are isomorphic to either $S_3$ or $\mathbb{Z}_6$.

Without knowing that, I was trying to derive how many structurally distinct groups of order 6 exists by constructing the multiplication tables.

And I came upon the following statement on this question:

Having all non-identity elements have order 2, means the group is abelian.

Is this trivial? Could I know that before trying to build the table?

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marked as duplicate by Matthew Towers, Community Sep 16 '16 at 14:52

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It's not very hard to show : if every element $x$ in a group is such that $x^2=e$, then for any $a,b$ $$a(ba)b=(ab)(ab)=ee=(aa)(bb)=a(ab)b,$$ and simplyfying $a$ and $b$ on the sides gives $ba=ab$.

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