If $G$ is a non-abelian group of order 10, prove that $G$ has five elements of order 2. I'm trying to prove this statement:

If $G$ is a non-abelian group of order $10$, prove that $G$ has five elements of order $2$. 

I know that if $a\in G$ such that $a\neq e$, then as a consequence of Lagrange's theorem $|a|\in \{2,5,10\}$. The order of $a$ cannot equal $10$, since then $G$ would be cyclic, and thus abelian which is a contradiction. Now this means that $|a|=2 $ or $|a|=5$. I know from  this question that $G$ has a subgroup of order $5$. This subgroup $H$ has prime order, so it is cylic, and all of its non-identity elements have order $5$. Now I need to show that the elements not in $H$ have order $2$. This is where I'm stuck.
I've tried assuming that an element $b \notin H$ has order $5$, in order to derive a contradiction, but to no avail. 
I also know from a previous exercise that if $G$ has order $10$, then it has at least one subgroup of order $2$, so I tried to assume toward a contradiction that $G$ has two subgroups of order $5$, and one subgroup of order $2$. I was trying to show that this would make $G$ abelian, but I couldn't. 
Any ideas?

Use *Sylow's theorems: 
The number of subgroups of order $5$ is congruent to $1$ modulo $5$, and a divisor of $\frac{10}5=2$. As there can't be $6$ since $G$ would have at least $25$ elements, there's only one subgroup of order $5$, say $H$, which is therefore a normal subgroup of $G$, generated by any of its elements $g$ distinct from $e$.
Let $a$ be an  element of order $2$;   its conjugate $gag^{-1}$ also has order $2$, and it is distinct from $a$, since this would mean $ga=ag$. The same would be true for any power of $g$, so that $G$ would be abelian.
Thus there are $5$ elements of order $2$.
