Let $n \ge 3$. Let $D_n = \langle r,s \rangle$, for $r^n=s^2=1$ and $rs=sr^{-1}$. Then $D_n$ (not $D_{2n}$) is the dihedral group of order $2n$.
Show that if $n$ is odd, then $D_n$ doesn't have a (EDIT: )proper normal subgroup of order $2m$, where $m$ divides $n$.
Question: Is there a way to go about doing this without necessarily knowing all the subgroups of $D_n$ and in particular that when $n$ is odd the only normal subgroups are (cyclic) subgroups of the cyclic subgroup $\langle r \rangle $?
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Or is there a way to show that when $n$ is odd the only normal subgroups are (cyclic) subgroups of the cyclic subgroup $\langle r \rangle $ without necessarily knowing all the subgroups of $D_n$?
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Perhaps normal subgroup implies non-intersection with the reflection half $\{s,sr,…,sr^{n-1}\}$?
If everything is allowed, then I'll just pretend I know all the subgroups of $D_n$ and that when $n$ is odd the only subgroups are (cyclic) subgroups of $\langle r \rangle $ :
Step 1. Assume/Prove the above (which is a BIG thing to assume. see keith conrad. There's this simpler version here.)
Step 2. For $order(r)=n$ and for any $t$ that divides $n$, we have $order(r^t)=\frac{n}{gcd(n,t)}=\frac{n}{t}$.
Step 3. Suppose on the contrary that $D_n$ has a normal subgroup $N$ of order $2m$. By step 1, $N$ is a (cyclic) subgroup of the cyclic subgroup $\langle r \rangle$.
Step 4: By Step 3, $N = \langle r^d \rangle$, for some $d$ that divides $n$.
Step 5. By Step 2, $order(r^d)=2m$ if and only if $2md=n$.
Step 6. Since n is odd, we have by step 5 that $order(r^d)$ can't be even.
Step 7. Therefore, steps 3-4 and 6 give our required contradiction.
Possibly relevant things:
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Previously, I attempted to exhibit that whether $n$ is odd or even, $D_n$ has subgroups of orders $m$ and $2m$: Prove dihedral group has subgroup of order $m$ and then of order $2m$, where $m$ divides $n$.
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Let $N$ be a normal subgroup of a group $G$. Let $H$ be a subgroup of $G$. Suppose both the order of $N$ and the index $[G:H]$ are finite and then coprime. Show that $N \subseteq H$.
Best Answer
John Smith Kyon, your idea is very good. If we remove all unnecessary things from your reasoning, we get a solution to our problem. It looks something like this.
Let $n$ be odd and let $N$ be a normal subgroup in $D_n$ of order $2m$.
Step 0. We know that $I=\{s,sr,\ldots,r^{n-1}\}$ is a complete list of all elements of order $2$ of group $D_n$.
Step 1. Since $N$ is of even order, we get $N\cap I\neq\varnothing$. Let $sr^k\in N$.
Step 2. Since $N$ is a normal subgroup of $D_n$, then $$ r^t(sr^k)r^{-t}=sr^{k-2t}\in N \hbox{ for all $t\in\mathbb{Z}$}. $$
Step 3. Since $n$ is odd, we obtain that the set $$ \{k-2t\mid t=0,1,\ldots,n-1\} $$ is a complete residue system modulo $n$. It follows that $I\subset N$. So $N=D_n$.