Show that $ G/G_m$ and $G/G_n$ are isomorphic groups. 
Let $G$ be the multiplicative group of complex numbers of modulus
  $1$ and $G_n,n\in \Bbb N$  the subgroup consisting of the $n$-th
  roots of unity. For positive integers $m$ and $n$, show that $ G/G_m$ and
  $G/G_n$ are isomorphic groups.

Below is my attempt at the problem:
Consider the mapping $\phi:G\to G^m$ where $G^m=\{z^m:z\in G\}$.
Then $G/G_m\cong G^m$  since $\ker \phi =G_m$.
Similarly $G/G_n\cong G^n$
If now I can show that $G^m$ and $G^n$ are isomorphic then we are done.
But the problem is how to show that $G^m\cong G^n$ ?
Please help.
 A: Consider $f:G/G_m \to G$ defined by $f(\overline g) = g^m$.
It is trivial to see that it is surjective, where by "trivial" I mean "a standard exercise in complex analysis".
To see that it is injective, let $f(\overline g) = f(\overline h)$, i.e. $g^m = h^m$. Then, $(gh^{-1})^m = 1$, so $gh^{-1} \in G_m$, so $\overline g = \overline h$.
Therefore, $f$ is an isomorphism, and we have $G/G_m \cong G \cong G/G_n$.
A: Define $\phi:G\rightarrow G$ by, $\phi(g)=g^n$ for all $g\in G$.
Easy checking: $\phi$ is group homomorphism and $\operatorname{Ker}(\phi)=G_n$.
In order to apply first isomorphism theorem to get the desired result we need to prove $\phi$ is a surjection.
For a given $g_0\in G$, let $g=e^{\frac1n\operatorname{Log}g_0}$, where $\operatorname{Log}(g_0)$ is the principal branch of $\operatorname{log}(g_0)$. Then $\phi(g)=g_0$.
Only remaining job is to show $g\in G$, that is, $|g|=1$.
Now, $|g|=|e^{\frac1n\operatorname{Log}g_0}|=|e^x|$, where $x=Re(\frac1n\operatorname{Log}g_0)=\frac1n\operatorname{log}|g_0|=0$, since $|g_0|=1$. Thus $|g|=1$, which proves it.
