How to prove that $\mathbb{C}_n$ is a subgroup of $(\Bbb C\setminus\{0\},\cdot)$. [closed]

I'm not sure what you're saying when you write things like $C_1 = C_{1+4n}$, but this is false. The cyclic group $C_n$ is a group with $n$ elements in it. Therefore, $C_n \neq C_m$ whenever $n \neq m$.

You also said "a subgroup is when its closed under the operation," but this is false. For example, the set $\mathbb Z$ of integers is a group with respect to addition, and $\mathbb N = \{0, 1, 2, ...\}$ is a subset of $\mathbb Z$ which is closed under addition, but it is not a subgroup.

Here is the correct definition of subgroup: if $G$ is a group, then a set $H$ is a subgroup of $G$ if the following conditions are met:

• $H$ is a subset of $G$

• $H$ has at least one element

• (Closure under the group operation from $G$) If $x$ and $y$ are elements of $H$, then so is $x \cdot y$.

• (Closure under the inverse operation from $G$) If $x$ is an element of $H$, then so is $x^{-1}$.

So to show that $C_n = \{ z \in \mathbb C : z^n = 1 \}$ is a subgroup of $\mathbb C^{\ast} = \mathbb C \backslash \{0\}$, you need to verify all the above. The first two conditions are obvious: clearly, $C_n$ is a subset of $\mathbb C^{\ast}$, almost by definition. And clearly $C_n$ is not the empty set. What should you do next?

Subgroup of A means that it is a group and subset of A at the same time using the same operation.

So you need to show that (i) $\mathbb{C_n}\subset \mathbb{C}-\{0\}$ and (ii) $\mathbb{C}_n$ is a group using the same operation as in $\mathbb{C}-\{0\}$.