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Theorem: the following exhausts all possible subgroups of $D_n$:

$1$. $\langle r^d \rangle$, where $d\mid n$

$2$. $\langle r^d,r^is \rangle$, where $d\mid n$ and $0\leq i \leq d-1$

It is easy to see why both of these would be subgroups. If $H\leq \langle r \rangle$, then since $\langle r \rangle$ is cyclic, $H = \langle r^d \rangle$ for some $d\mid n$. I'm having issues proving that if $H$ contains some element of the form $r^ms$ (it's not a subgroup of $\langle r \rangle$) then it is necessarily of the second type.

I have found a proof in here (Theorem 3.1), yet I was merely wondering if there was a shorter way of proving the result.

Shaun
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Sam
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    I don't think that the proof given in Keith Conrad's notes can be improved. It can be differently presented perhaps, according to your taste or needs, but this wouldn't change the essential arguments. – Dietrich Burde Nov 22 '19 at 19:59
  • @DietrichBurde do you disagree with the posted answer of Leo? it removes a lot of the conjugacy index stuff – BCLC Oct 21 '21 at 18:16

1 Answers1

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I thought to give my question another try today. For what it's worth, the following seems somewhat shorter and makes use of more elementary notions:

We'll consdier two different types of subgroups. Those that contain reflections (elements of the form $r^xs$), and those that do not.

If $H \leq D_n$ doesn't contain a reflection, then it is a subgroup of $\langle r \rangle$, and is thus equal to $\langle r^d \rangle$ where $d|n$.

Suppose $H \leq D_n$ does contain a reflection, then since $1=r^n \in H$, $H \cap \langle r \rangle \neq \emptyset$. Let $d$ be the smallest positive number such that $r^d \in H$. If $r^e \in H-\langle r^d \rangle$ and $e=dk+f$ for $f>0$ then $r^f \in H$, which contradicts $d$ being minimal. Thus $H=\langle r^d \rangle \cup \{ \text{reflections} \}$. Let $r^xs \in H$ so that $x$ is again minimal. We can see $H$ contains

$$\{ r^{dk},r^{dk+x}s:0\leq k \leq \frac{n}{d}-1 \}$$ and if $H$ contains another reflection $r^ys$ with $0\leq y<n$, then $r^yss^{-1}r^{-x} = r^{y-x} \in \langle r^d \rangle$ implies $y=dh+x$. Therefore $h=\frac{y-x}{d}<\frac{n}{d}$ and

$$H=\{ r^{dk},r^{dk+x}s:0\leq k \leq \frac {n}{d}-1 \}$$.

One can then use the same argument as Konrad to show this set is equal to $\langle r^d,r^is \rangle$ where $d|n$ & $0\leq i \leq d-1$. If $\langle r^d, r^is \rangle = \langle r^e, r^js \rangle$ then by the argument presented before $d=e$, so all such sets are different. Furthermore $\langle r^d \rangle \neq \langle r^e, r^js \rangle$ since only the second one contains a reflection, so the two lists of subgroups are disjoint.

Sam
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