Determine number of subgroups of index $2$ in $D_n$. Question:
Determine the  number of subgroups of index $2$ in $D_{2n}$.
Consider $D_{2n}=\langle r,s|r^n=1, s^2=1, sr=r^{-1}s\rangle$.
Take $N=\langle r\rangle$ and consider canonical homomorphism $G$ to $G/N$.
Then $\langle s\rangle$ is isomorphic to $G/N$.  Hence $D_n$ is semidirect
product of $C_2$ and $C_n$.
So my claim is there is unique subgroups $N$ of index 2.
Thanks
my argument correct?  If so how do I prove the claim.
 A: *

*You write: "Then $\langle s\rangle$ is isomorphic to $G/N$." well, that's true. But, why?

*Yes, $D_n$ is the semidirect product as you state.

*So, your "claim" is rather a "guess" so far: you only showed one subgroup of index $2$. So, we are at "there is at least one". And the guess is mostly correct, a counterexample is $n=4$ (the 8 element dihedron grp).

*For your guess, you should also prove that there is no other.

A: An index $2$ subgroup is always normal, so counting index two subgroups is the same as counting normal index $2$ subgroups, which is the same as counting nontrivial homomorphisms to the two-element group $\{ \pm 1\}$. Since $D_{2n}$ is generated by $r$ and $s$, such a homomorphism is determined by where it sends $r$ and $s$, and there are a priori three possibilities: $(r,s)\mapsto (1,-1)$, $(-1,1)$, or $(-1,-1)$. If $n$ is odd then the relation $r^n=1$ requires $r$ to be sent to $1$, so in fact there is only one subgroup of index $2$, namely $\langle r \rangle$. If however $n$ is even then all three possibilities respect all the relations, and so there are three index-$2$ subgroups, namely $\langle a \rangle$, $\langle a^2,b\rangle$ and $\langle a^2, ab\rangle$.
