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...
Here's an alternative solution without using Sylow's theorems. 

Note that $|G|\equiv 2 \pmod 4 \Rightarrow |G|=2(2k+1)$ for some $k\in \mathbb{Z}$. 

Since $2$ divides $|G|$, by Cauchy's theorem there exists an element $g\in G$ of order $2$. Now $\langle g\rangle$ is a subgroup of order $2$. Now since $G$ is abelian we have that $\langle g\rangle $ is a normal subgroup, and hence $G/\langle g\rangle $ is a group. Now by Lagrange's theorem $|G/\langle g\rangle | = 2k+1$, which is odd.

Suppose there was another element $h\in G$ with order $2$ and $h\neq g$. Then we have that $ h + \langle g \rangle $ is an element of order $2$ in $G/\langle g\rangle$. Thus $\langle h + \langle g\rangle \rangle $ is a subgroup of $G/\langle g\rangle$ or order $2$. However, this is a contradiction since $|G/\langle g\rangle|$ is odd.
Hint: $|G|$ is $2(1+2p)$, a finite abelian group is a product of groups isomorphic to $\mathbb{Z}/p^l$ where $p$ is a prime this implies that the component associated to $2$ is $\mathbb{Z}/2$.

https://en.wikipedia.org/wiki/Finitely_generated_abelian_group#Classification