# Proving that $\langle r \rangle$ is the only normal cyclic subgroup of $D_{2n}$ of index $2$

I've been doing a few exercises about the dihedral group, and a while ago I tried to solve this one:

Show that $$\langle r \rangle$$ is the only normal cyclic subgroup of $$D_{2n}$$ of index $$2$$.

Here $$D_{2n}$$ denotes the dihedral group of order $$2n$$ and $$r$$ denotes the rotation of order $$n$$. I will denote a reflection by $$b$$.

Now, I think I managed to solve it for $$n \geq 3$$, but I think the exercise is false for $$n = 2$$. Indeed, the following subgroups of $$D_4$$ are all distinct, normal and cyclic and of index $$2$$: $$\{1, r\}, \{1, b\}, \{1, br\}$$. Right? So I think my professor should have added the restriction $$n \neq 2$$.

My attempt at a proof: notice that a generator of a cyclic subgroup of order $$2$$ in $$D_{2n}$$ cannot be in $$D_{2n} - \langle r \rangle$$. Now, for all $$i \in \{1, 2, \cdots, n - 1\}$$, we have that $$\langle r^{i} \rangle$$ is contained in $$\langle r \rangle$$, therefore the only cyclic subgroup of order $$2$$ of $$D_{2n}$$ is $$\langle r \rangle$$. Indeed, if $$\langle r^{i} \rangle$$, with $$2 \leq i \leq n - 1$$ has index $$2$$, then $$| \langle r^{i} \rangle |= n$$, but since $$\langle r^{i} \rangle \subset \langle r \rangle$$ and $$\operatorname{ord}(r) = n$$ we have that $$\langle r^{i} \rangle = \langle r \rangle$$, proving uniqueness. Therefore the only cyclic subgroup of index $$2$$ in $$D_{2n}$$ is $$\langle r \rangle$$. In particular, the only normal cyclic subgroup of index $$2$$ in $$D_{2n}$$ is $$\langle r \rangle$$.

Am I correct?

• Somehow you are supposed to put proof-verification in the tags, not in the title... – WhatsUp Nov 18 '19 at 0:17
• @WhatsUp fixed. – Matheus Andrade Nov 18 '19 at 0:18
• It's easy to see that $\langle r \rangle$ is the only cyclic group of index two, i.e. order $n$, as a reflection has order 2, and a rotation generates a subgroup of $\langle r \rangle$. The normal part comes for free.. – Dzoooks Nov 18 '19 at 0:20
• @Dzoooks that was my idea too. But the exercise is false for $D_4$, right? – Matheus Andrade Nov 18 '19 at 0:21
• @MatheusAndrade Your comment that is false for $n=2$ (and true for $n\geq 3$) is correct. Whether this is a mistake depends on what definition is used for the dihedral group. Some authors include $n\geq 3$ as part of the definition, especially if it comes from the geometric side (defining is as the group of symmetries of a regular polygon, for example). – verret Nov 18 '19 at 0:38

Note: I denote a reflection by $$s$$ instead of $$b$$.
Since $$r$$ and $$s$$ generate D$$_{2n}$$, we need only consider combinations of $$r$$ and $$s$$. We have $$|s|$$ = $$2$$ and $$|r|$$ = $$n$$, so $$$$ is one possible cyclic subgroup. To prove that there exist no other cyclic subgroups of order $$n$$, consider the following:
Any element in D$$_{2n}$$ can be written in the form of either sr$$^i$$ or r$$^i$$, where $$0$$ $$\le$$ i $$\le$$ n.
Then $$||$$ is the least positive integer $$k$$ such that $$(sr^i)^k$$ = 1.
Note that $$sr^i$$ * $$sr^i$$ = $$s^2$$*1 = 1, using the relation $$rs$$ = $$sr^{-1}$$. Hence the order of any cyclic subgroup of the form <$$sr^i$$> is $$2$$ as the order the generator is 2. Since you have specified $$n$$ $$\ge$$ $$3$$, we must have our cyclic subgroup as order $$n$$ since the index will be $$2n$$ $$/$$ $$||$$.
Hence since we considered all possible cyclic subgroups of $$D_{2n}$$, we can conclude that our previously found cyclic subgroup, $$$$ is the only cyclic subgroup of order $$n$$. It is normal since all index 2 subgroups are normal in any group $$G$$, as given H a subgroup of G with index 2, we have two cosets aH and H. Then H is both a left coset and a right coset so aH = Ha and so H is normal.