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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?

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  • $\begingroup$ Somehow you are supposed to put proof-verification in the tags, not in the title... $\endgroup$
    – WhatsUp
    Nov 18, 2019 at 0:17
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    $\begingroup$ @WhatsUp fixed. $\endgroup$ Nov 18, 2019 at 0:18
  • $\begingroup$ 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.. $\endgroup$
    – Dzoooks
    Nov 18, 2019 at 0:20
  • $\begingroup$ @Dzoooks that was my idea too. But the exercise is false for $D_4$, right? $\endgroup$ Nov 18, 2019 at 0:21
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    $\begingroup$ @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). $\endgroup$
    – verret
    Nov 18, 2019 at 0:38

1 Answer 1

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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 $<r>$ 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 $|<sr^i>|$ 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 $2n$ $/$ $|<sr^i>|$.

Hence since we considered all possible cyclic subgroups of $D_{2n}$, we can conclude that our previously found cyclic subgroup, $<r>$ 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.

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