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).