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.

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

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.


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