An element of order $n$ generates a normal subgroup of $D_n$

Question

Let $a$ be an element of order $n$ of $D_n$. Show that $\langle a\rangle \lhd D_n$ and $D_n/\langle a\rangle \cong \mathbb Z_2$.

Proof: Let $K = <a>$ for some a ∈ G. Let H ≤ K be an arbitrary subgroup. Since $H ≤ K = <a>$ it follows that $H = <a^d>$ for some integer d. If |a| = 1 then a = 1, H = K = {1} and, obviously, $H \lhd G$.

Is what I have right? I think to prove the second part I use Lagrange but I'm not sure how.

Answer 1

If $D_n$ is the dihedral group, then it has $2n$ elements. It is a general fact that every subgroup of order $n$ is normal.