# Prove order of a group is even

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.

• why in the last step you have $\phi(r)=\phi(r)^{-1}$? – mathpadawan Nov 18 '18 at 2:36
• Because I assumed $\varphi(s) = 1$ – Kiarash Jamali Nov 18 '18 at 2:38

## 1 Answer

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.