A group with finitely generated normal subgroup and finitely generated quotient is finitely generated itself

Question

Let $G$ be a group with $G \trianglerighteq N$ normal subgroup. Assume that $N$ is finitely generated, and $G /N$ (the quotient group) is finitely generated as well. Is $G$ finitely generated?

I think that the answer is no, and I wanted to use the following example: $G = \mathbb{Q}$, $N=\mathbb{Z}$. My only remaining question is, how to show that $\mathbb{Q} /\mathbb{Z}$ is finitely generated?(if at all).

And if it isn't, any other ideas?

Answer 1

Hint: if $G/N=\langle \bar{g_1}, \bar{g_2}, \cdots, \bar{g_k} \rangle$, then certainly $G=\langle g_1, g_2, \cdots, g_k \rangle N$. So if in addition $N$ is finitely generated, what can you conclude about $G$?

Answer 2

I think $\mathbb{Q}/\mathbb{Z}$ is not finitely generated abelian group.

Let $\mathbb{Q}/\mathbb{Z}=<q_1+\mathbb{Z},...,q_n+\mathbb{Z}>$ then we ca find $a_1,...,a_n\in\mathbb{Z}$ and $b\in\mathbb{N}$ st $q_i=\dfrac{a_i}{b}$. Now if $p$ is a prime st $p\not| b$ then $\dfrac{1}{p}+\mathbb{Z}\not\in \mathbb{Q}/\mathbb{Z}$ contradiction!

(If $\dfrac{1}{p}+\mathbb{Z}\in\mathbb{Q}/\mathbb{Z}$ then $\dfrac{1}{p}+\mathbb{Z}=l_1\dfrac{a_1}{b}+...+l_n\dfrac{a_n}{b}+\mathbb{Z}\Rightarrow p|b$)

So for the initial question as @Nicky Hekster says

Let $g\in G$ then $g+N\in G/N$ so $g+N=a_1g_1+N+...+a_ng_n+N=(a_1g_1+...+a_ng_n)+N$ but $N$ is finitely generated so ...

$g-a_1g_1+...+a_ng_n=m\in N=<q_1,...,q_k>\Rightarrow g= \sum a_ig_i+\sum b_j q_j $