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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.


my argument correct? If so how do I prove the claim.

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You haven't made an argument. You've said a few things about $D_{2n}$, and then made an unjustified (and false) assertion about the answer to your question. – Chris Eagle Nov 1 '12 at 17:52
sorry for that... – user38764 Nov 1 '12 at 17:56

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 $\langle a^2, ab\rangle$.

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  1. You write: "Then $\langle s\rangle$ is isomorphic to $G/N$." well, that's true. But, why?
  2. Yes, $D_n$ is the semidirect product as you state.
  3. 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).
  4. For your guess, you should also prove that there is no other.
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This is not fair, giving an answer for an 0 accpt rate here and then have a minus vote??? I fix it Berci. :) – Babak S. Nov 24 '12 at 13:08

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