I am trying to prove that an abelian group of order 6 has exactly one element of order 2.
I know there is at least one by Cauchy's Theorem, so I am trying to show there are no more than one by contradiction.
Suppose there are $a,b, a \neq b$ such that $a^2 = b^2 = e$. Then also $ab$ has order $2$, so we have two remaining elements (aside from $e, a, b, ab$) of which at least one of them, say $c$, has order $3$ by Cauchy. Then $ac$ has order $6$, but also $bc$ has order $6$ and there is only one element of order 6 (since $e$ has order 1, $a, b, ab$ have order 2, and $c$ has order 3) so $ac = bc$ and therefore $a = b$. This is a contradiction.
Is this proof correct? Is there a "better" proof? One without Cauchy's Theorem?