For the first part see here. I will follow the following notation:

**Notation. ** with and

**Lemma 3**. Given let For every let Let Then is a subgroup of and We also have

*Proof*. If for some then and thus because Therefore because So Clearly and because Thus Proving that is a subgroup of is easy. Just note that every element of is the inverse of itself (because they all have order two) and also note that for all because Finally, the set has elements because clearly if and only if if and only if

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

*Proof*. Suppose that is a subgroup of There are two cases to consider.

*Case 1* . By Lemma 1, the number of these subgroups is

*Case 2* . In this case, by Lemma 2, we have and for some Let Since is a subgroup of which is a cyclic group of order we have

Let and be as they were defined in Lemma 3. Now, since is not contained in there exists some such that Then, since is a subgroup, we must have for all Thus and so and therefore, by we have Thus, since we must have The converse is obvously true, i.e. given and is a subgroup of by Lemma 3, and because it contains So the subgroups in this case are exactly the ones in the form where and Thus, by Lemma 3, the number of subgroups in this case is

So, by case 1 and case 2, the number of subgroups of is

Note that we didn’t just find the number of subgroups of We also found all the subgroups.

**Example**. Find all subgroups of

**Solution**. There are subgroups. Four of them are obtained from case 1 in the proof of the theorem. They are the subgroups of Since the subgroups in this case are and There are subgroups left and they are in the form where and So or Also, by the proof of the last part of Lemma 3, iff So those subgroups are:

Note that