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?

Hint:
It has one single subgroup $\;P\;$ of order two and one single subgroup of order five $\;Q\;$ iff $\;P,Q\lhd G\;$ , but then $\;G=PQ=\ldots\;$ 
