Vrftjkry

I am trying to solve this question and wanted to know whether my proof was correct.

Suppose that $n geq 3$, $n$ is odd, $G$ is a non-trivial group and $varphi : D_{2n} rightarrow G$ is a surjective homomorphism.

(a) Prove that $|G|$ is even.
(b) Prove that every proper normal subgroup of $G$ has odd order.

My attempt for a: Since $G$ is not trivial and is equal to $varphi(D_{2n})$, then either $varphi(s) not = 1$ or $varphi(r) not = 1$. If $varphi(s) not = 1$, then we have $varphi(s)^2 = 1$ and we have found an element of order 2 in $G$ so it must be even. If $varphi(r) not = 1, varphi(s) = 1$, then we have that $varphi(sr) = varphi(r^{-1}s) Rightarrow varphi(s)varphi(r) = varphi(r)^{-1}varphi(s) Rightarrow varphi(r) = varphi(r)^{-1} Rightarrow varphi(r)^2 = 1$ and since $varphi(r) not = 1$, we have again found an element of order 2 in $G$.

My attempt for b: I'm not sure about this one, but I first note that by the first isomorphism theorem, $G cong D_{2n}/ker(varphi)$. Any proper normal subgroup of $G$ now has to be isomorphic to one of $D_{2n}/ker(varphi)$. Then, by the fourth isomorphism theorem, it has to be isomorphic to a normal subgroup of $D_{2n}$. Now I don't know how to proceed.

Now take any proper normal subgroup $H$ of $D_{2n}$. What you have proved is that $D_{2n}/H$ has even order. say $2m$.

Now $|D_{2n}|=|D_{2n}/H||H|=2n Rightarrow 2m|H|=2n Rightarrow m|H|=n$

Since $n$ is odd, $|H|$ is also odd. So any proper normal subgroups of $D_{2n}$ have odd order if $n$ is odd.