## Subgroups of dihedral groups (1)

Posted: February 17, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , , , ,

The goal is to find all subgroups of $D_{2n},$ the dihedral group of order $2n.$

Definition. Let $n \geq 1$ be an integer. The number of divisors of $n$ is denoted by $\tau(n).$ Also the sum of divisors of $n$ is denoted by $\sigma(n).$ For example, $\sigma(8)=1 + 2+4+8=15$ and $\tau(8)=4.$

We have the following cute result and we will prove it in the second part of our discussion.

Theorem. (Stephan A. Cavior, 1975) If $n \geq 3,$ then the number of subgroups of $D_{2n}$ is $\tau(n) + \sigma(n).$

The theorem says that the number of “all” subgroups, including $\{1\}$ and $D_{2n},$ is $\tau(n) + \sigma(n).$

Lemma 1. The number of subgroups of a cyclic group of order $n \geq 1$ is $\tau(n).$

Proof. Let $G$ be a cyclic group of order $n.$ Then $G \cong \mathbb{Z}/n \mathbb{Z}.$ A subgroup of $\mathbb{Z}/n \mathbb{Z}$ is in the form $d \mathbb{Z}/n \mathbb{Z}$ where $d \mathbb{Z} \supseteq n \mathbb{Z}.$ The condition $d \mathbb{Z} \supseteq n \mathbb{Z}$ is obviously equivalent to $d \mid n. \ \Box$

Lemma 2. Let $b$ be an element of order $n$ in $D_{2n}$ and let $H$ be any subgroup of $D_{2n}.$  Then either $H \subseteq \langle b \rangle$ or $|H \cap \langle b \rangle| =d$ and $|H|=2d$ for some $d \mid n.$

Proof. Let $N=\langle b \rangle.$ Clearly $N$ is a normal subgroup of $D_{2n}$ because $[D_{2n}:N]=2.$ Thus $HN$ is a subgroup of $D_{2n}$ and hence

$|HN| \mid 2n. \ \ \ \ \ \ \ \ \ \ \ \ (*)$

On the other hand,

$\displaystyle |HN|=\frac{|H| \cdot |N|}{|H \cap N|}=\frac{n |H|}{|H \cap N|}.$

Therefore, by $(*),$ $\frac{|H|}{|H \cap N|} \mid 2.$ Hence either $|H| = |H \cap N|$ or $|H|=2|H \cap N|.$ If $|H|=|H \cap N|,$ then $H = H \cap N$ and thus $H \subseteq N.$ If $|H|=2|H \cap N|,$ then let $|H \cap N|=d$ and so $|H|=2d.$ Clearly $d \mid n$ because $H \cap N$ is a subgroup of $N$ and $|N|=n. \ \Box$

In part (2) we will prove Cavior’s theorem and also we will find all subgroups of $D_{2n}$ explicitly.