The goal is to find all subgroups of the dihedral group of order

**Definition**. Let be an integer. The number of divisors of is denoted by Also the sum of divisors of is denoted by For example, and

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

**Theorem**. (Stephan A. Cavior, 1975) If then the number of subgroups of is

The theorem says that the number of “all” subgroups, including and is

**Lemma 1**. The number of subgroups of a cyclic group of order is

*Proof*. Let be a cyclic group of order Then A subgroup of is in the form where The condition is obviously equivalent to

**Lemma 2**. Let be an element of order in and let be any subgroup of Then either or and for some

*Proof*. Let Clearly is a normal subgroup of because Thus is a subgroup of and hence

On the other hand,

Therefore, by Hence either or If then and thus If then let and so Clearly because is a subgroup of and

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

[…] Update: I have found an group-theoretic answer at https://ysharifi.wordpress.com/2011/02/17/subgroups-of-dihedral-groups-1/ […]

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