# If $H\leq G$, for which $n$ is $f_n: H/nH\to G/nG$ an isomorphism?

Here is a question from an old prelim I am having trouble with..

Suppose $G$ is an abelian group and $H\leq G$ such that $[G:H]=m<\infty$. For any $n\geq 1$ define the map $H\to G/nG$ by composing the inclusion $H\to G$ and the natural projection $G\to G/nG$. Since $nH\leq nG$, this induces a homomorphism $f_n:H/nH\to G/nG$ given by $f_n(h\mod nH)=h\mod nG$.

(1) Show that if $\gcd(n,m)=1$, then $f_n$ is an isomorphism.

(2) Show that if $\gcd(n,m)>1$, then $f_n$ is not surjective.

For (1), my plan is to show that $|\ker f_n|$ divides both $n$ and $m$ and therefore must be $1$ by the gcd condition. Then we would have $f_n$ is an injective homomorphism between two groups of size $n$, hence an isomorphism.

$|\ker f_n|$ certainly divides $n$ by Lagrange since it is a subgroup of $H/nH$ which has order $n$.

I feel like I am getting messed up with defining the kernel of this map...

$\ker f_n=\{h+nH:h+nG=nG\}=\{h+nH:h\in nG\}=\{ng+nH:g\in G\}$

Is this third definition of the kernel OK? If so how can I show that $|\ker f_n|$ divides $m=[G:H]$? I know that $[nG:nH]=m$ as well. I guess this will be useful but I don't see how.

For (2) we just need to find, for example (I'm guessing this would be easiest), a left coset of $nH$ that is not itself $nH$ which maps to the identity $nG\in G/nG$. I feel like a better understanding of (1) would help me to find such an element for (2).

-
Sorry, why does $H/nH$ have order $n$? –  Dylan Moreland Aug 17 '11 at 21:55
I think your expressions for the kernel are all fine. I'm worried that $\operatorname{Ker} f_n$ could be infinite, but can't think of a good example. –  Dylan Moreland Aug 17 '11 at 23:23
Good point. I guess I was using my intuition about the integers mod $n$. Is it possible for $G/nG$ to be infinite for an abelian group $G$? –  RHP Aug 18 '11 at 1:46
I think so. What about $G = \prod_{n = 1}^\infty \mathbf Z$? –  Dylan Moreland Aug 18 '11 at 2:05
@RHP: It certainly can. $H/nH$ is an abelian group of exponent $n$, but it need not be of order $n$. –  Arturo Magidin Aug 18 '11 at 3:30

1. If $n, m$ are coprime, then there exist $a, b \in \mathbf Z$ such that $an + bm = 1$. So you can write each element $x \in G$ as $$x = (an)x + (bm)x.$$ As Rolando says, it's probably better to look at the kernel of the map $H \to G/nG$. Now, if $y$ is in the kernel then $y = nx$ for some $x \in G$. It would be great if we had $x \in H$. I claim that this follows from the above expression and the fact that $m$ is an exponent for the image of $x$ in $G/H$. If you do that, proving surjectivity should be clear.

2. Let $p$ be a prime dividing both $n$ and $m$. It seems easier to show that in this situation, multiplication by $n$ is not surjective on $G/H$. Indeed, this is a homomorphism and by Cauchy's theorem there is an element of $G/H$ having period $p$.

-

1) Some hints :

you have a map : $H\longrightarrow G/nG$ given by the composition of the inclusion $H\rightarrow G$ and the surjection $G\rightarrow G/nG$. What is the kernel of the map ?

You should then know that the induced map : $H/nH\longrightarrow G/nG$ is injective. The hypothesis on $m$ and $n$ is related with the surjectivity of the map : if $|G/H|$ has order coprime to $n$, what can you say on the order of the elements of $H$?

-