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?
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 $ba=ab$.