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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.

Shaun
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Nana
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4 Answers4

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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.

HJRW
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    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 06 '10 at 06:36
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    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
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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 $h_{ij}$.

Nuno
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  • Thanks Nuno, but I'll need help on showing the that H is generated by all of H_ij. Thanks. – Nana Dec 05 '10 at 09:08
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    @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 05 '10 at 09:22
  • @John: Follow Arturo's instructions. Let me know if you could finish it. @Arturo Magidin: Thanks. – Nuno Dec 05 '10 at 14:58
  • @John @ Arturo. Thanks very much for your help. I think I got it now. – Nana Dec 05 '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 01 '14 at 20:19
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    @mez It can be proved this $t$ is exactly $t_1=1$ – Danielsen Aug 14 '14 at 08:47
  • Restating Arturo's comment using equations: $h=g_{i_1}...g_{i_{\alpha}}$ (repeats allowed) for some arbitrary $\alpha\in\mathbb{N}$. Then $h=g_{i_1}g_{i_2}...g_{i_{\alpha-1}}g_{i_{\alpha}}=(h_{1i_{1}}t_{k_{1i_{1}}})g_{i_2}...g_{i_{\alpha-1}}g_{i_{\alpha}}=h_{1i_{1}}(t_{k_{1i_{1}}}g_{i_2})...g_{i_{\alpha-1}}g_{i_{\alpha}}=...=h_{1i_{1}}...h_{1i_{\alpha}}t$. As stated by Danielsen, we know that $t=1$. Hence, all elements of $H$ are generated by ${h_{ij}:i,j}$. – J.G.131 May 20 '24 at 18:10
  • Why does $t=1$? Proof by contradiction gives a sufficient explanation...though to put it explicitly, if $t\neq1$, then $h=h_0t\in Ht$ where $h_0=h_{1i_{1}}...h_{1i_{\alpha}}$. But $H\cap Ht=0$? With this contradiction, we deduce that in fact $t=1$. – J.G.131 May 20 '24 at 18:19
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If $G$ is finitely generated by $A\subset G$, equipping $G$ with the word metric, we can look at $G$ as a proper metric space $(G,d_{A})$. Then $H$ acts continuously (by isometries) on $(G,d_{A})$. Provoking Švarc–Milnor lemma, we get that $H$ is f.g and quasi-isometric to $G$ (looks like $G$ looking at both $G$ and $H$ from far distance).

user1729
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Hussain Rashed
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    +1, although I once had a lecturer who talked about using a nuclear bomb to blow up a bridge. This has a similar feeling... – user1729 Oct 10 '19 at 15:19
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    @user1729: The part of the Svarc—Milnor lemma needed — that a group acting properly discontinuously on a proper path-connected space is finitely generated — is pretty easy. It’s far from a nuclear bomb. To put the difficulty of the Svarc—Milnor lemma in context, I think Pierre de la Harpe calls it the “fundamental observation of geometric group theory”. I’ve always felt a little ashamed that the fundamental result of geometric group theory is just an observation! – HJRW Aug 10 '20 at 13:02
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    @HJRW Ha, yes, it's not hard and certainly not nuclear! My point, I believe, was that the use of quasi-isometries themselves was nuclear - their use requires a different and (from the OPs point of view) probably unexpected way of thinking. Which is not a bad thing, and I had upvoted the answer when I posted the comment :-) [For some reason, I don't think your covering-spaces argument is nuclear, but rather "intuitive".] – user1729 Aug 10 '20 at 18:19
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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.

Olod
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