How to prove that $\mathbb{C}_n$ is a subgroup of $(\Bbb C\setminus\{0\},\cdot)$. $\forall n \in \mathbb{N} \setminus \{0\}$  prove that $\mathbb{C}_n=\{z \in \mathbb{C}\;|\;z^n=1\}$ is a subgroup of $(\mathbb{C}\backslash\{0\},\cdotp)$ and that is a cyclic group of order $n$.
I know that a subgroup is when it's closed under the operation. Then to try to use the definition I noticed that $\mathbb{C}_1 = \mathbb{C}_{1+4n}$, $\mathbb{C}_2 = \mathbb{C}_{2+4n}$, $\mathbb{C}_3= \mathbb{C}_{3+4n}$, $\mathbb{C}_4 = \mathbb{C}_{4+4n}$, but I'm not sure if this is correct or not because I tried to prove it using induction but I failed. With that proven I thought that I could study the four sets one by one and see if they are subgroups to therefore prove that $\mathbb{C}_n$ is subgroup.
 A: 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?
A: 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 $\mathbb{C}-\{0\}$.
A: To show : for fixed $n\in \mathbb{N}$ $H= (\Bbb C_n,\cdot )\le G=( \Bbb C\setminus \{0\}, \cdot)  $
One step subgroup  test : $G$ be a group and $\emptyset \ne H\subseteq G $ if $\forall a, b \in H $ implies $ab^{-1} \in H$ , then $H\le G$.
Now, Clearly $H$ is non empty as $1\in H \space \space [1^n =1] $
Choose, $z, w\in H $ then $z^n=1 \text{ and } w^n =1 $
$w\in H \implies w\in G $ as $G$ is a group
$w^{-1} \in G $ and $w^{-1}=\frac{1}{w}$
And, $(zw^{-1})^n = \frac{z^n}{w^n}= 1 $
Hence, $zw^{-1}\in H$ and $H\le G$
Now we claim that $H$ is a finite cyclic group of order $n$.
First try to understand the set
$\Bbb C_n=\{z\in \Bbb C : z^n =1 \}$
$z^n =1 $
Suppose, $z=re^{i\theta}$
$[re^{i\theta}]^n = 1$
$r^n e^{i n \theta}=1$ [using De Moivre's formula]
Hence, $r=1 $ and $ n \theta =2ki \pi$ for $k=0,1,...,n-1$
Hence, $\{z\in \Bbb C : z^n = 1 \}= \{e^{\frac {2ki\pi}{n}}{ : k =0,1,...,n-1 }\}$
Then, $H=\Bbb C_n=\langle e^{\frac{2 i \pi}{n}}\rangle$
A: Since $\Bbb C_n$ is a finite, nonempty subset of $\Bbb C\setminus \{0\}$, it suffices to show that $\Bbb C_n$ is closed. Let $w,z\in\Bbb C_n$. Then $w^n=1=z^n$. Thus $(wz)^n=w^nz^n=1$ (because, say, $w,z$ commute). Hence $wz\in\Bbb C_n$. Hence $\Bbb C_n\le\Bbb C\setminus\{0\}$.
A generator of $\Bbb C_n$ is $e^{\frac{2\pi i}{n}}$.

Also, $\Bbb C_i\neq \Bbb C_{i+4n}$.

NB: The elements of $\Bbb C_n$ are known as the "$n$th roots of unity".
