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:

  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 $N \subseteq H$.

  • 1
    $\begingroup$ @Leo $D_n$ not $D_{2n}$. The order of $D_n$ is $2n$ not $n$ $\endgroup$
    – BCLC
    Oct 21, 2021 at 19:21
  • $\begingroup$ @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 $D_n$ $\endgroup$
    – BCLC
    Oct 21, 2021 at 19:25
  • 2
    $\begingroup$ Oh I was completely lost after all. $\endgroup$
    – Sam
    Oct 21, 2021 at 19:26
    – BCLC
    Oct 21, 2021 at 19:38
  • 1
    $\begingroup$ 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 $G$? $\endgroup$
    – Derek Holt
    Oct 21, 2021 at 19:53

3 Answers 3


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 $N=D_n$.

  • $\begingroup$ 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. $\endgroup$
    – BCLC
    Oct 26, 2021 at 14:32
  • $\begingroup$ kabenyuk, is this right also please? Prove dihedral group has subgroup of order m and then of order 2m, where m divides n. $\endgroup$
    – 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.

  • $\begingroup$ 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. $\endgroup$
    – 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


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 $N = D_n$.



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