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?

$G$ \ $N$ the set $G-N$ or do you mean $G/N$ quotient group? — Nicky Hekster, Aug 25 '18 at 12:07

score 5 · Accepted Answer · answered Aug 25 '18 at 12:22

5

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 $G$?

answered Aug 25 '18 at 12:22

Nicky Hekster

49,281

That $G=\langle g_1, g_2, \cdots, g_k \rangle \langle n_1,n_2, \cdots, n_r \rangle$? – ChikChak Aug 25 '18 at 12:26
1

Yes well done! Hence $G$ is f.g. $G=\langle g_1, ..., g_k, n_1, ..., n_r \rangle$. – Nicky Hekster Aug 25 '18 at 12:36
The explanation why $G/N=\langle \bar{g_1}, \bar{g_2}, \cdots, \bar{g_k} \rangle$ implies $G=\langle g_1, g_2, \cdots, g_k \rangle N$ is that $G$ is the union of the cosets of $N$?(i.e., $G = \cup G / N = \langle g_1, g_2, \cdots, g_k \rangle N$?) – ChikChak Aug 25 '18 at 13:55
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@ChikChak: if $g \in G$, then in $G/N$, $\bar{g}=\bar{x_1}^{s_1} \cdot \bar{x_2}^{s_2} \cdots \bar{x_r}^{s_r}$, with $\bar{x_i} \in {\bar{g_1}, \bar{g_2}, \cdots , \bar{g_k}}$ and $s_i \in \mathbb{Z}$. So, in $G$, $g={x_1}^{s_1} \cdots {x_r}^{s_r} \cdot n$ for a certain $n \in N$. And $n$ can also be written as a finite product of the finite number of generators of $N$. Got it? – Nicky Hekster Aug 25 '18 at 15:38
Nicky Hekster please help here: finitely generated quotient group: From $G/H = \langle g_1H, ..., g_m H \rangle$, we have $G=\langle g_1, ..., g_m \rangle H$? @ChikChak – BCLC Oct 21 '21 at 14:27
Nicky Hekster and @ChikChak are you sure we can just combine the indices in product sets like $\langle g_1, ..., g_m \rangle \langle h_1, ..., h_n \rangle = \langle g_1, ..., g_m, h_1, ..., h_n \rangle$ ?!?! – BCLC Oct 21 '21 at 14:32
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@John Smith Kyon yes this is true. Can you prove it? Not really difficult. Try to do it by showing two containments. – Nicky Hekster Oct 21 '21 at 15:05
Nicky Hekster thanks for replying! i'll try now. but what about in $S_3$ like $\langle$ (1 2)$\rangle$ $\langle$ (1 3)$\rangle$ is not equal to $\langle$ (1 2), (1 3) $\rangle$ or something? – BCLC Oct 21 '21 at 15:09
Nicky Hekster wait got it. in this case the LHS = $\langle g_1, ..., g_m \rangle \langle h_1, ..., h_n \rangle$ is given to be a subgroup of $G$ (actually the whole $G$) and that's why we can just combine the indices? – BCLC Oct 21 '21 at 15:16
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Yes you got it! – Nicky Hekster Oct 21 '21 at 15:21
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Or one of the factors in normal. If not then you gave the smallest counterexample. Well done! – Nicky Hekster Oct 21 '21 at 15:25
Thank you very very much Nicky Hekster! Btw, I posted an answer just now: (internal) product of cyclic subgroups or generated subgroups in general: $\langle x \rangle$ $\langle y \rangle$ = $\langle x,y \rangle$? – BCLC Oct 21 '21 at 15:44

1123581321 · Answer 2 · 2018-08-25T12:33:30.290

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 $

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

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