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Let $G=\langle e, r,..., r^{n-1},s,sr,...,sr^{n-1} \rangle $ be a dihedral group with $2n$ elements, for $ 3 \leq n$. Prove that the only normal subgroups of $G$ are $\langle r^d \rangle$ (where $d$ divides n) and $G$ itself in the case that $n$ is odd.


So here is what I did:

Suppose that $sr^i \in H \unlhd G$. Then $(sr^j)(sr^i)(sr^j)^{-1}=sr^{i-2j} \in H$

Now I need to show that from $sr^{i-2j} \in H$ it follows that $sr^i \in H$ for all $i$, but I don't know how to do that. I think I need to use the fact that $n$ is odd but I just don't see it.

If I can show this I can conclude that if a normal subgroup contains $sr^i$, then $H=G$ must be the case.

For the second part I need to prove that $\langle r^d \rangle$ is a normal subgroup of $G$ when $d$ divides $n$:

$r^i r^d (r^i)^{-1}=r^d \in \langle r^d \rangle$ for all $i$

and also $sr^i r^d (sr^i)^{-1}=s^2 r^{-d}=(r^d)^{-1} \in \langle r^d \rangle$ for all $i$. Since we checked for the generators we can conclude that $\langle r^d \rangle$ is a normal subgroup of $G$. But I don't see why this isn't the case if $d$ doesn't divide $n$. Nowhere in my proof did I encounter that.

So I actually have three questions:

  • How do I show that if $sr^i$ is an element of some normal subgroup of G that that subgroup needs to equal G.
  • Did I make a mistake in the second part? Is it necessary for $d$ to divide $n$ and if so where did I go wrong in my proof?
  • How do I conclude that a normal subgroup has to be one of these two? Can I just say that these are the only possible subgroups?

Thanks in advance!

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Note on your assumption is sr^i in H. In another,   sr^k are elements of order 2, in particular sr ^ {i-2j} is sr ^ {i}; and subgroups of order 2 are {e, sr^i} (Sylow subgroups becvaus n is odd). Therefore they are conjugated to each other, which means that if H contains a sr^i it contains all sr^i.

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