Problem about subgroup of $D_n$ Prove that every subgroup of $D_n$ , either every member of subgroup is a rotation or exactly half of them are rotations.
Intuitively, if every member is a rotation then they will form a subgroup because we can rotate them as much as we like (closure) and other properties will also be satisfied. But how do we prove that exactly half of them will be rotations form a subgroup. Please give Hints to start! 
Thanks

This is what the question is really asking you:
Let $D_n$ be the dihedral group of order $2n$ and $R$ the order $n$ group of rotations.
The if $H$ is any subgroup of $D_n$, either $[H:H \cap R] = 1$ or $[H:H \cap R] = 2$.
If $H \subseteq R$, then $H \cap R = H$, and we have $[H:H \cap R] = [H:H] = 1$.
Otherwise, we can invoke (via the second isomorphism theorem):
$[H: H \cap R] = [HR:R]$.
Now $HR$ is a subgroup of $D_n$ properly containing $R$, and $[D_n:R] = 2$, so $HR = D_n$. Can you continue?
