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

There are only 2 groups of order 10, namely the cyclic group and the dihedral group of symmetries of a regular pentagon. The reflections in the dihedral group give you the five desired elements of order $2$.

Now, to prove that there are only 2 groups of order 10, let $a,b$ be elements of orders $2,5$ respectively. Consider the elements $1$, $b$, $b^2$, $b^3$, $b^4$, $a$, $ab$, $ab^2$, $ab^3$, and $ab^4$, and notice that they are all distinct. To determine the group, it suffices to determine what $ba$ is. This is the same as determining what $a^{-1}ba$ is. By nonabelianness, we know that $a^{-1}ba \neq b$, so merely check against all other elements of the group...

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 $\;G=PQ=\ldots\;$