Isomorphism between two factor groups, where group $G$ consists of all the numbers with magnitude 1 
Let $G$ be the multiplicative group of complex numbers of modulus $1$ and $G_n$ is the subgroup of $G$ consisting of the $n$-th roots of unity. For $m,n \in \mathbb N$, show that $G/G_n$ and $G/G_m$ are isomorphic.


My approach.
Let us define a function $\phi_n $ : $ G \to G$, such that   $\phi_n $($g_1$) $ =g_1^n$, $ \forall g_1 \in G$.
Clearly $G_n$ is the kernel of the homomorphism $\phi_n $. Let us say $H_n , H_m$ are the range of $\phi_n $ and $\phi_m $ respectively.
Now let us define another function $ \pi $ : $H_n \to H_m$, such that $h_i \mapsto h_i$. Now essentially for all $h_i$, there exist atleast one $g_i$ such that $g_i^n=h_i$.
Now ofcourse $\forall $ $g_i \in G$,     $g_i^{\frac{n}{m}}$ $\in G$, as both of them are of unit magnitude. This implies $\forall $ $g_i \in G$, $g_i^n \in G/G_m$.
This implies the function $\pi$ is valid and of course one-one.
Interchanging the role of $ m$ and $n$, we can say there exists an one-one function between $H_m$ and $H_n$ as well.
So, $H_n$ and $H_m$ are isomorphic groups, so of course $G/G_n$ is isomorphic to  $G/G_m$  .
Is it all right?
 A: Note that $G\simeq\Bbb R/2\pi\Bbb Z$ via the exponential map $\Bbb R\to \Bbb C:x\mapsto e^{ix}$. 
Via the same map you have $G/G_n\simeq \Bbb R/\frac{2\pi}{n}\Bbb Z$.
Finally, the map $\Bbb R/\frac{2\pi}{n}\Bbb Z\to \Bbb R/ 2\pi\Bbb Z: x\mapsto nx$ is an isomorphism.
Therefore $G/G_n\simeq G$ independently of $n$.

Notes about your own proof: 


*

*note that $H_n$ and $H_m$ are both $G$ itself since $\phi_n$ is surjective for all $n$ (every unit complex number has a unit $n$-th root). Therefore $\pi$ is the identity map $G\to G$. I'm not sure you're going anywhere with this map.

*what is the purpose of the subscripts $1$ in $g_1$ and $i$ in $g_i$? (as far as I can see there is none and you can drop them)

*what is the meaning of $g^{\frac nm}$ when $g\in G$? Any unit complex number has $m$ $m$-th roots.

*when you say 

"This implies $\forall $ $g_i \in G$, $g_i^n \in G/G_m$."

That doesn't make sense. If $g\in G$, you can consider the class of $g_i^n$ in the quotient $G/G_m$, but it doesn't intrinsically live there! 


*

*the existence of an injective function $H_n\to H_m$ and another injective map $H_m\to H_n$ is not a proof that $H_n$ and $H_m$ are isomorphic. What you need to find is one map that is bijective and such that both the map and its inverse are homomorphisms.

