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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?

Mikasa
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L--
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1 Answers1

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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.
Arnaud Mortier
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  • Yes, I understood your argument but can you please tell me whether my method is right or not? – L-- Jul 13 '18 at 14:49
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    @ArnabChowdhury I was in the process of editing my answer. Done now. – Arnaud Mortier Jul 13 '18 at 14:52
  • Well, I have something to say about your last paragraph. I mean suppose there exists $f_1: A \to B$ and $f_2 : B \to A$ where both are injective. Of course range($f_1$) is isomorphic to $A$ and range($f_2$) is isomorphic to $B$. This implies $A$ contains an isomorphic copy of $B$, and $B$ contains an isomorphic copy of $A$; These are true together iff $A$ and $B$ are isomorphic to each other. This is how I deduced it. Although it is very much possible that I am missing something. @Arnaud Mortier – L-- Jul 14 '18 at 03:57
  • @ArnabChowdhury This might very well be true, but then you need a reference or a proof, and you also need to state that you are using that, because it is neither obvious nor classic. – Arnaud Mortier Jul 14 '18 at 05:55
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    @ArnabChowdhury It is, in fact, wrong: see edit. – Arnaud Mortier Jul 14 '18 at 05:57