Show that dihedral group of twice odd order doesn't have a normal subgroup of order $2m$, where $m$ divides $n$. 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 $?

*

*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 $D_n$?


*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:

*

*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 $n$.


*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 $N \subseteq H$.
 A: 
is there a way to show that when  is odd the only normal [proper] subgroups
are (cyclic) subgroups of the cyclic subgroup ⟨⟩ without necessarily
knowing all the subgroups of $D_n$?

Yes-- just check the  character table of $D_n$ (where $n$ is odd).  There are two $1$-dim representations -- the trivial representation and the sign / determinant representation and neither will work.  Using the induced representation / Frobenius Reciprocity to build off cyclic subgroup $H$ generated by $r$, all other representations are 2 dimensional and identically 0 on the 2nd coset $aH$.
Normal subgroups are in the kernel of some homomorphism, so the character table tells you that any proper normal subgroup cannot include the second coset $aH$.
In general you can extract an awful lot amount of information from the character table of a group despite not knowing much about the actual group.  And induced representations are relatively straightforward when dealing with index 2 subgroups.
A: 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 $N=D_n$.
A: Edit: Didn't care anymore after the mistake of my instructor about the proper normal subgroup. I just answered with saying that proper normal subgroup doesn't intersect the reflections. And then did the rest of the stuff.

Wait I think I got it. Let's prove that a normal subgroup $N$ with order $2m$ while $n$ is odd can't intersect the reflections, i.e. we can't have $N \cap \{s,sr,...,sr^{n-1}\} \ne \emptyset$:
Step 1. Suppose on the contrary $N$ intersects the reflections, i.e. $N \cap \{s,sr,...,sr^{n-1}\} \ne \emptyset$. Then $sr^k \in N$, for some $k=0,1,...,m-1$.
Step 2. Because $N$ is normal subgroup, we have that $N$ contains the elements $a(sr^k)a^{-1}$ with $a = s, r$.

*

*$sr^{n-k} = (s)(sr^k)(s)^{-1}$

*$r^{k-2}s = (r)(sr^k)(r)^{-1}$
Step 3. Then because $N$ is a regular subgroup, we have that $N$ contains
$r^{k-2}ssr^{n-k}=r^{k-2}r^{n-k}=r^{n-2}$.
Step 4. Now for $order(r)=n$, we have $order(r^{n-2})=\frac{n}{\gcd(n,n-2)}$. Because $n$ is odd (and because $n \ge 3$), we have that $\gcd(n,n-2)=1$. Hence, $order(r^{n-2})=\frac n 1 = n$.
Step 5. By Steps 3-4, we have I think a contradiction with Lagrange's theorem...Or well if this doesn't work just say that all of $\langle r \rangle$ is a subset of $N$. Combined with the $sr^k \in N$, we have the rest of the reflections in $N$ because $sr^i=sr^k(r^{i-k})$. Therefore $D_n \subseteq N$. Hmmm...the exact contradiction is...well it's a contradiction if $m \ne n$.
As for $m = n$, we have $N = D_n$.
QED!
