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.