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I have a finitely generated group $G$, and a (normal) subgroup $N<G$. I want to show that if N has finite index in G, then N is finitely generated. This is not a duplicate, as I am interested in understanding the proof below and not some other proof that might have already been given in another thread.

So let $G=<g_1,...,g_k>$ be finitely generated. Since $N$ has finite index in G we have $G=a_1N\cup...\cup a_lN$ for some $a_1,...,a_l\in G$. This means that for every $i$ I can find some $j_i$ s.t. $g_i\in a_{j_i}N$, i.e. $h_i:=a_{j_i}^{-1}g_i \in N$. We then have $H=<h_1,...,h_k>\subset N$.

Now the argument is that (i) $H$ has finite index in $G$ and (ii) that therefore it has finite index in $N$.

Pick finitely many representatives of the cosets from $H/N:$ $b_1,...,b_m$, then $N=<h_1,...,h_k,b_1,...,b_m>$

I understand the last point, but not the two listed as (i) and (ii). Could someone elaborate on why those are true?

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  • $\begingroup$ I'll stick my neck out and second your confusion on part (i). Note that (ii) follows from (i) because index is multiplicative. I see that the generators of $G$ are included in $\bigcup_i a_{j_i} H$, but I don't see how to use this to prove that $H$ has finite index in $G$. $\endgroup$ – Dustan Levenstein Nov 5 '16 at 14:38
  • $\begingroup$ With index being multiplicative you mean that once we know that $H$ is finite in $G$, we have $(N:H)=\frac{(G:H)}{(G:N)}$? $\endgroup$ – azureai Nov 5 '16 at 15:08
  • $\begingroup$ Yes. More precisely, if $\{a_i\}_{i \in I}$ is a selection of coset representatives of $H$ in $N$ (possibly infinite), and $\{b_j\}_{j \in J}$ are coset representatives of $N$ in $G$, then $\{a_i b_j\}_{(i, j) \in I \times J}$ form a system of coset representatives of $H$ in $G$. $\endgroup$ – Dustan Levenstein Nov 5 '16 at 15:20
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    $\begingroup$ The proof of (i) should use the normality somehow since, as far as I can tell, it isn't used anywhere else.... $\endgroup$ – Mike F Nov 5 '16 at 16:11

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