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.

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Comments
  1. a2c2a says:

    That was a useful post. Also if n is prime then the number of subgroups is n + 3, right?

    Proof:

    Number of factors of a prime = 2 (1 and the prime itself, n)
    Sum of factors of a prime = 1 + the prime itself, n
    2 + 1 + the prime itself = 3 + n

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