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?