Quotient group isomophic to $\mathbb{Z}$

Question

In a certain document I am asked to show that if, given a group $G$, there is a normal subgroup $N$ such that $G/N \simeq \mathbb{Z}$, $\forall n \in \mathbb{N}$ there exists a subgroup $H$ such that $|G/H| = n$. I have no idea about how to start. I was wondering if I could choose a subgroup of $N$ such that $|N/H| = n$ and it would be over maybe, but I even don't know if that subgroup would neccesarily be normal. Can somebody give me just a hint?
By the way, I was also asking myself if, given a group $G$ such that there is an onto homomorphism between $G$ and $\mathbb{Z}$, one could conclude that $|G| = |\mathbb{Z}|$;

Answer 1

$N$ can be arbitrary (take $G = N \oplus \mathbb Z$), so we can't prove anything about it.

Let $\phi$ be isomorphism $G / N \to \mathbb Z$. Define homomorphism $\psi: G \to \mathbb Z_n$ as $\psi(x) = \phi(x) \mod n$ (can you show that it is homomorphism?). Take $H$ as kernel of $\psi$.

For your second question - no, any group can be a quotient of an arbitrary large group. For example, take $G = \mathbb R \oplus \mathbb Z$, and $N = \mathbb R \oplus \{0\}$.

Answer 2

To answer the first question, use the duality between normal subgroups and homomorphic images given by the first isomorphism theorem: $$ G/\ker(\phi)\cong\phi(G) $$

So, working with homomorphic images, our starting assumption is that $G\twoheadrightarrow\mathbb{Z}$. Writing $\mathbb{Z}_n$ for the cyclic group of order $n$, we know that $\mathbb{Z}\twoheadrightarrow\mathbb{Z}_n$, and so we may compose these homomorphisms to get a surjection $\phi:G\twoheadrightarrow\mathbb{Z}_n$. By the first isomorphism theorem, $\ker(\phi)$ is a normal subgroup such that $G/\ker(\phi)\cong\mathbb{Z}_n$, and hence $\ker(\phi)$ had index $n$ in $G$, as required.

For the second question, just take $G=H\times\mathbb{Z}$ for any group $H$ with cardinality $|H|>|\mathbb{Z}|$.