# subgroups of finitely generated groups with a finite index

Let $G$ be a finitely generated group and $H$ a subgroup of $G$. If the index of $H$ in $G$ is finite, show that $H$ is also finitely generated.

• – Ofir Dec 5 '10 at 7:17

Hint: Suppose $G$ has generators $g_1, \ldots, g_n$. We can assume that the inverse of each generator is a generator. Now let $Ht_1, \ldots, Ht_m$ be all right cosets, with $t_1 = 1$. For all $i,j$, there is $h_{ij} \in H$ with $t_i g_j = h_{ij} t_{{k}_{ij}}$, for some $t_{{k}_{ij}}$. It's not hard to prove that $H$ is generated by all the $h_{ij}$.

• Thanks Nuno, but I'll need help on showing the that H is generated by all of H_ij. Thanks. – Nana Dec 5 '10 at 9:08
• @John: Take an arbitrary $h\in H$. Write it as a product of $g_j$, $g_{i_1}\cdots g_{i_k}$. Then $g_{i_1} = t_1g_{i_1}$ can be written as $h_{1i_1}t_{k_{1i_1}}$. Now look at $t_{k_{1i_1}}g_{i_2}$ and replace it with the product of an $h_{rs}$ times a $t$; then look at that $t$ times $g_{i_3}$, etc. – Arturo Magidin Dec 5 '10 at 9:22
• @John: Follow Arturo's instructions. Let me know if you could finish it. @Arturo Magidin: Thanks. – Nuno Dec 5 '10 at 14:58
• @John @ Arturo. Thanks very much for your help. I think I got it now. – Nana Dec 5 '10 at 16:02
• @ArturoMagidin But then at last we can write $h$ as a product of $H_{ij}$ with some $t$. How to get rid of the $t$? – mez May 1 '14 at 20:19

Here's the topological argument. The fact that $G$ is finitely generated means that $G=\pi_1(K)$ for $K$ a CW-complex with finite 1-skeleton. Let $\widehat{K}$ be the covering space corresponding to $H$. Then $H=\pi_1(\widehat{K})$, and $\widehat{K}$ also has finite 1-skeleton, so $H$ is finitely generated.

• This is a nice way of thinking about it. The proof that I had in mind was sort of halfway between this and the purely algebraic arguments others are giving. Namely, it is enough to show this for free groups, and this reduced to some easy considerations on covering spaces of finite graphs. – Pete L. Clark Dec 6 '10 at 6:36
• Pete - yes, that's essentially the same proof. The fact that every group is a quotient of a free group is just the assertion that every complex has a one-skeleton. – HJRW Dec 12 '10 at 15:21

Well, the standard argument is as follows.

Let $$g \mapsto [g] \qquad (g \in G)$$ be a function which is constant on all right cosets of $H,$ and we require $$[e]=e.$$

It is easy to see that $$u [u]^{-1} \in H, \quad [[u]]=[u], \quad [[u]v]=[uv] \qquad (u,v \in G) \qquad \qquad (*)$$ Now let $$S = \{ [g] : g \in G\}$$ and $Y=Y^{-1}$ be a symmetric generating set of $G.$ Then the set $$\{ s y [sy]^{-1} : s \in S, y \in Y\}$$ is a generating set of $H$ (a finite one, if both $S$ and $Y$ are finite $\iff$ the index of $H$ in $G$ is finite and $G$ is finitely generated).

For suppose that a product $y_1 \ldots y_r$ is in $H$ where $y_k \in Y$ ($k=1,\ldots,r$). Let, for example's sake, $r=3.$ Then $$y_1 y_2 y_3 = y_1 [y_1]^{-1} \cdot [y_1] y_2 [[y_1] y_2]^{-1} \cdot [[y_1] y_2] y_3 [[[y_1] y_2] y_3]^{-1} \qquad \qquad (**)$$ where in the right hand side we have a product of elements of $H$ by (*), since $$[[[y_1] y_2] y_3]=[y_1 y_2 y_3]=e;$$ the same $(*)$ also simplifies the right hand side of $(**)$ as $$y_1 [y_1]^{-1} \cdot [y_1] y_2 [y_1 y_2]^{-1} \cdot [y_1 y_2] y_3 [y_1 y_2 y_3]^{-1}=y_1 y_2 y_3 [y_1 y_2 y_3]^{-1}=y_1 y_2 y_3$$ Now the induction step in general must be easy.

• Thanks for giving me the standard approach. I appreciate it. – Nana Dec 5 '10 at 16:52