the number of subgroups of the unit group $\mathbb{Z}_{n}^{\times}$

Question

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!

Answer 1

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 $(\mathbb{Z}/2020)^\times$.

Answer 2

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.