Show that $R_{n}=\left\{z \in \mathbb{C} \mid z^{n}=1\right\}$ is a subgroup of $S=\{z \in \mathbb{C} : |z|=1\}$. Show that for each $n \in \mathbb{N}, n \geq 2$ the set of roots $n$ n-th of the unity: $R_{n}=\left\{z \in \mathbb{C} \mid z^{n}=1\right\}$ is a subgroup of $S=\{z \in \mathbb{C}  : |z|=1\}$.
I try:
$1 \in R_n $ because $1^n=1$ and $1*z^n=1*1=1$ for all $z \in R_n$
If $z,x \in R_n$ then $z^n*x^n=(z*x)^n=1*1=1 $ then $z^n*x^n \in R_n$
and if $z\in R_n$ then $z^{-1} \in R_n$ because $z^{-1}=\frac{1}{z^n}=\frac{1}{1}=1=z*z^{-1}$
is right?
 A: I will use the one-step subgroup test.
Fix $n\in\Bbb N$, $n\ge 2$.
Clearly $1\in R_n\neq\varnothing $ and $R_n\subseteq S$. Indeed,
$$\begin{align}
z^n=1&\implies|z^n|=1\\
&\implies|z|^n=1\\
&\implies|z|=\pm1\\
&\implies|z|=1.
\end{align}$$
Let $y,z\in R_n$. Then $y=e^{i2k\pi/n}, z=e^{i2\ell\pi/n}$ for $k,\ell\in \Bbb Z$. So
$$\begin{align}
yz^{-1}&=e^{i2k\pi/n}e^{-i2\ell\pi/n}\\
&=e^{i2(k-\ell)\pi/n}\\
&\in R_n
\end{align}$$
since $k-\ell\in\Bbb Z$.
Hence $R_n\le S$.
A: You seem to confuse the definition of $R_n$ being about elements $z$ so that $z^n$ has a property, with believing the elements of $S$ are the $z^n$.  Any you seem to confusing your task of discovering things about elements $x,z \in R_n$ with discovering things about $x^n,z^n$.

*

*In proving that $1$ is an identity of $R_n$ you must prove for all $z \in R_n$ that $1*z = z$.  You seem to think we want to prove that $1\cdot z^n = z^n$ (or $1\cdot z^n = 1$.... your argument is not clear.

Anyway as $1$ is the identity element for $\mathbb C$, $1$ will inherit its identity function for any subset containing $1$ (whether or not the subset is a group or not).  So all we need to do is show $1 \in \mathbb R_n$.  Which it is as $1^n = 1$.


*And in proving muliplication is closed within elements of $R_n$, we must show if $x,z\in R_n$ that $xz\in R_n$.  Start to show this correctly but then you conclude with the wrong conclusion $x^nz^n \in R_n$ (for which your argument isn't valid).

What we wish to show is that $x,z\in R_n$ will imply $xz \in R_n$.  And, as multiplication is commutative $(xz)^n = x^nz^n$ and as $x,z\in R_n$ we hae $x^nz^n=1\cdot 1 =1$ and so $xz\in R_n$.
Your third argument is incomprehensible.
We must show that if $z\in R_n$ that $z^{-1}\in R_n$.  That is if $z^n =1$ then $(z^{-1})^n = 1$.  You could do it the way it seemed like you were attempting by noting as $z^n =1$ then $z^{-1} = z^{-1}z^n = z^{n-1}$ and $(z^{n-1})^n = z^{n(n-1)} = (z^n)^{n-1} = 1^{n-1} =1$ so $z^{-1}\in R_n$ but that seems a bit convoluted.  There is a more straightforward way:  $(z^{-1})^n = (z^{-1})^n\cdot 1 = (z^{-1})^n\cdot z^n = 1$.
....
But before you do any of this you must actually prove $R_n$ is a subset of $S$.  That is if $z^n = 1$ you must prove $|z| = 1$.
