Show that if $G$ is abelian, and $|G| \equiv2 \mod 4$, then the number of elements of order $2$ in $G$ is $1$. I've tried proving it by contradiction, assuming the number of elements is different than one, which, by Sylows $3$, implies that $|G|=2^xm$ with $m$ odd. With that I managed to show that $m\equiv1\mod4$, but kinda got stuck there...

In fact $|G|\equiv 2\pmod 4$ implies that
$$|G|=2m$$
with $m$ odd (if it were are higher power of $2$, it would be congruent to $0$ instead). So  Sylow $2$-subgroups are of order $2$. You should be able to take it from there.
