the number of subgroups of the unit group $\mathbb{Z}_{n}^{\times}$ Is there a way to find the number of subgroups of the multiplicative group $\mathbb{Z}_{n}^{\times}$?
For example, the number of subgroups of $\mathbb{Z}_{2020}^{\times}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{4}\times\mathbb{Z}_{100}$.
When the given number $n$ become quite large, it was hard to find it because there were more and more cases to think about.
Anyone can help my problem? or Give some advice. Thank you!
 A: Not really answering your question, but we have $\text{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong \mathbb{Z}/n\mathbb{Z}^{\times}$. Maybe you are able to find an article or anything in that way. Otherwise I would try the following:
Let $n = p_1^{\nu_1} \cdot \dots p_r^{\nu_r}$ be the prime decomposition of $n$. Then we get $\mathbb{Z}/n\mathbb{Z}^{\times} \cong \mathbb{Z}/p_1^{\nu_1}\mathbb{Z}^{\times} \times \dots \times \mathbb{Z}/p_r^{\nu_r}\mathbb{Z}^{\times}$ and for prime powers we have 
$\mathbb{Z}/p^{\nu}\mathbb{Z}^{\times} \cong \begin{cases}
               C_1               & p = 2 \;\text{and}\; \nu = 1\\
               C_2 \times C_{2^{\nu -2}}               & p = 2 \;\text{and}\; \nu \geq 2\\
               C_{\phi({p^\nu})} = C_{p^{\nu - 1}(p - 1)} & \text{otherwise}
           \end{cases}$
By that you can at least get a grasp on the subgroups of the factors of $\mathbb{Z}/n\mathbb{Z}^{\times}$ and therefore also of the subgroups of $\mathbb{Z}/n\mathbb{Z}^{\times}$ arising as a product, but I guess that is probably what you have been doing. 
A: Yes, it is a complicated mess in general because you have to count the number of "diagonals".
First, rewrite the abelian group $G$ into primary decomposition $G=G_{p_1}\oplus G_{p_2}\oplus\dots\oplus G_{p_r}$.  Then the subgroup lattice of $G$ is the product of the lattice of $G_{p_i}$, so in particular the number of subgroups is the product for various prime factors of $\lvert G\rvert$.  In our present example $(\mathbb{Z}/2020)^\times=(C_2\oplus C_4\oplus C_4)\oplus C_{25}$, and it is clear $C_{25}$ has 3 subgroups.
To determine the number of subgroups in the $2$-part, we will need to get our hands slightly dirty.  First a lemma

Lemma (Goursat): For every subgroup $L$ of $H\oplus K$, there is a natural isomorphism
  $$
\frac{(L+K)\cap H}{L\cap H}\cong\frac{(L+H)\cap K}{L\cap K}
$$
  Conversely, let $U\leq L_H\leq H$ and $V\leq L_K\leq K$ be subgroups of the direct summands.  For every isomorphism $\phi\colon L_H/U\to L_K/V$, there exists a subgroup $L\leq H\oplus K$ such that $L_H=(L+K)\cap H$, $L_K=(L+H)\cap K$, $U=L\cap H$, $V=L\cap K$, namely
  $$
L=\{x+y\mid x\in L_H, y\in\phi(x+U)\}.
$$

Let's define a "diagonal" $D$ with respect to the decomposition $G=H\oplus K$ to be a subgroup of $G$ such that $D+H=D+K=G$ and $D\cap H=D\cap G=\{0\}$.  Then there is a bijection between diagonals with respect to $G=H\oplus K$ and isomorphisms $\phi\colon H\to K$, given by regarding $D$ a the graph of the isomorphism.  Moreover, for an arbitrary subgroup $L$ of $H\oplus K$, we have $L/((L\cap H)\oplus(L\cap K))$ is a diagonal of $\frac{\operatorname{proj}_HL}{L\cap H}\oplus\frac{\operatorname{proj}_KL}{L\cap K}$.
So we can count the number of subgroups that are not product of subgroups by counting diagonals in the product lattice.

Claim: $C_4\oplus C_4$ has 15 subgroups.

Proof This can be established by laborious counting, or establishing the 6 possible diagonals with respect to the subgroup lattice $\{0\}<C_2<C_4$ of the $C_4$s.  $2^2=4$ diagonals $C_2\to C_2$, and there is two diagonals $C_4\to C_4$.  QED.

Claim: There are 54 subgroups of $C_2\oplus C_4\oplus C_4$.

Proof: Using the previous claim, there are 30 subgroups that are direct factors $C_2\oplus(C_4\oplus C_4)$.  Now we need to count diagonals $C_2\to C_2$, which is the same as the number of inclusions of index 2 subgroups in the subgroup lattice of $C_4\oplus C_4$.  There are 12 from direct product, each $C_2\to C_2$ contributes 2, and finally the two $C_4\to C_4$ contributes two each from $\langle(2,2)\rangle$ to the subgroup and to $\langle(2,0),(1,1)\rangle$.  So there are a total of 24 diagonals, so 30+24=54 subgroups.  QED.
So in total, there are 108 subgroups of $(\mathbb{Z}/2020)^\times$.
