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

Notation. D_{2n} =\{a^rb^s : \ 0 \leq r < 2, \ 0 \leq s < n \}, with o(a)=2, \ o(b)=n and ab=b^{-1}a.

Lemma 3. Given d \mid n, let m = n/d. For every 0 \leq i < n let A(i,d) = \{ab^{i+km}: \ 0 \leq k < d \}. Let B(i,d) = A(i,d) \cup \langle b^m \rangle. Then B(i,d) is a subgroup of D_{2n} and |B(i,d)|=2d. We also have |\{B(i,d) : \ 0 \leq i < n \}|=m.

Proof. If ab^{i+km}=ab^{i+rm}, for some 0 \leq k,r < d, then b^{(k-r)m}=1 and thus d \mid k-r, because o(b)=n=md. Therefore k=r because 0 \leq k,r < d. So |A(i,d)|=d. Clearly A(i,d) \cap \langle b^m \rangle = \emptyset and |\langle b^m \rangle| = d, because o(b)=n. Thus |B(i,d)|=|A(i,d)| + |\langle b^m \rangle|=2d. Proving that B(i,d) is a subgroup of D_{2n} is easy. Just note that every element of A(i,d) is the inverse of itself (because they all have order two) and also note that ab^s = b^{-s}a, for all s, because ab=b^{-1}a. Finally, the set \{B(i,d) : \ 0 \leq i < n \} has m elements because clearly B(i,d)=B(j,d) if and only if A(i,d)=A(j,d) if and only if i \equiv j \mod m. \ \Box

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

Proof.  Suppose that H is a subgroup of D_{2n}. There are two cases to consider.

Case 1 . H \subseteq \langle b \rangle. By Lemma 1, the number of these subgroups is \tau(n).

Case 2 . H \nsubseteq \langle b \rangle. In this case, by  Lemma 2, we have |H|=2d and |H \cap \langle b \rangle|=d, for some d \mid n. Let n = md. Since H \cap \langle b \rangle is a subgroup of \langle b \rangle, which is a cyclic group of order n, we have

H \cap \langle b \rangle = \langle b^m \rangle. \ \ \ \ \ \ \ \ \ \ (*)

Let A(i,d) and B(i,d) be as they were defined in Lemma 3. Now, since H is not contained in \langle b \rangle, there exists some 0 \leq i < n such that ab^i \in H. Then, since H is a subgroup, we must have ab^ib^{km} \in H, for all k. Thus ab^{i + km} \in H and so A(i,d) \subseteq H and therefore, by (*), we have B(i,d) \subseteq H. Thus, since |H|=|B(i,d)|=2d, we must have H=B(i,d). The converse is obvously true, i.e. given d \mid n and 0 \leq i < n, B(i,d) is a subgroup of D_{2n}, by Lemma 3, and B(i,d) \nsubseteq \langle b \rangle because it contains A(i,d). So the subgroups in this case are exactly the ones in the form B(i,d), where 0 \leq i < n and d \mid n. Thus, by Lemma 3, the number of subgroups in this case is

\sum_{d \mid n} | \{B(i,d) : \ 0 \leq i < n \} | = \sum_{d \mid n} n/d = \sum_{d \mid n} d = \sigma(n).

So, by case 1 and case 2, the number of subgroups of D_{2n} is \tau(n) + \sigma(n). \ \Box

Note that we didn’t just find the number of subgroups of D_{2n}. We also found all the subgroups.

Example. Find all subgroups of D_{12}.

Solution. There are \tau(6)+\sigma(6)=4+12=16 subgroups. Four of them are obtained from case 1 in the proof of the theorem. They are the subgroups of \langle b \rangle. Since o(b)=6, the subgroups in this case are \{1\}, \ \langle b \rangle, \ \langle b^2 \rangle and \langle b^3 \rangle. There are 12 subgroups left and they are in the form B(i,d), where 0 \leq i < 6 and d \mid 6. So d = 1, 2,3 or 6. Also, by the proof of the last part of Lemma 3, B(i,d)=B(j,d) iff i \equiv j \mod 6/d. So those 12 subgroups are:

B(0,1), B(1,1),B(2,1),B(3,1),B(4,1),B(5,1), B(0,2),B(1,2),B(2,2),B(0,3),B(1,3),B(0,6).

Note that B(0,6)=D_{12}.

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