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 $$\{s,sr,...,sr^{n-1}\}$$?

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:

1. 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 $n$.

2. 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 $$N \subseteq H$$.

• @Leo $D_n$ not $D_{2n}$. The order of $D_n$ is $2n$ not $n$
– BCLC
Oct 21, 2021 at 19:21
• @Leo Please help in my answer. I try to show we can't have such a normal subgroup even though I don't know all the subgroups of $D_n$
– BCLC
Oct 21, 2021 at 19:25
• Oh I was completely lost after all.
– Sam
Oct 21, 2021 at 19:26
• @Leo WAIT NO. YOUR COMMENT HELPED ME REACH CONCLUSION. SEE MY UPDATED ANSWER. THANK YOU!!!!!!
– BCLC
Oct 21, 2021 at 19:38
• Firstly, your problem statement is technically incorrect. $D_n$ does have a normal subgroup of order $2m$ with $m$ dividing $n$, namely $D_n$ itself. So you need to specify proper normal subgroup in your problem statement. Secondly, do you know about the commutator subgroup $[G,G]$ (sometimes called the derived subgroup $G'$) of a group $G$? Oct 21, 2021 at 19:53

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 $$N=D_n$$.

• thank you very much kabenyuk! though i think i could've figured this out in time for homework submission had i not been so damn tired already figuring out the proper subgroup thing. but i guess that i could've figured it out (insert stuff i just said) is kinda the point of the 1st part of your answer.
– BCLC
Oct 26, 2021 at 14:32
• kabenyuk, is this right also please? Prove dihedral group has subgroup of order m and then of order 2m, where m divides n.
– BCLC
Oct 27, 2021 at 18:49

is there a way to show that when 𝑛 is odd the only normal [proper] subgroups are (cyclic) subgroups of the cyclic subgroup ⟨𝑟⟩ 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.

• thanks user8675309 but we don't really have character table, representations, etc. but well i guess i am indeed on the right track to show the only normal subgroups are cyclic subgroups of $\langle r \rangle$. Please help me out in my answer where I tried to do something along this path.
– BCLC
Oct 21, 2021 at 19:23

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 $$a = s, r$$.

1. $$sr^{n-k} = (s)(sr^k)(s)^{-1}$$
2. $$r^{k-2}s = (r)(sr^k)(r)^{-1}$$

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!