If $G$ is a group and $H \leq G $, such that $[G:H] =n$, and $H$ is finitely generated, then $G$ is finitely generated

Question

I have a task from my group theory course

If $G$ is a group and $H \leq G $, such that $[G:H] =n$, and $H$ is finitely generated, then $G$ is finitely generated.

I had an idea using $G = H \cup g_1H \cup \dots \cup g_{n-1}H$, and maybe showing that the union is finitely generated, but I think I'm missing something, or that I'm completely wrong.

Am I in the right way?

Thanks for the help!

Mod out by the core of $H$, then use that if $S$ generates $G$ modulo $N$ and $T$ generates $N$, then $S\cup T$ generate $G$. — Arturo Magidin, Nov 14 '21 at 19:56

score 2 · Accepted Answer · answered Nov 14 '21 at 19:49

Take a finite generating set $S$ of $H$. It is easy to show that $G=\langle g_1,...,g_{n-1}, S\rangle$. Indeed, since this subgroup contains $S$, it must contain $H$. And since it also contains the element $g_i$, it contains the coset $g_iH$. So this subgroup contains all the cosets, and so it is all $G$.

If $G$ is a group and $H \leq G $, such that $[G:H] =n$, and $H$ is finitely generated, then $G$ is finitely generated

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