The group $\mathrm{GL}(n,\mathbb{Z})$ is finitely generated: take for example diagonal matrices, permutations and one elementary matrix (upper triangular). Are there some simple / nice examples of non-finitely generated subgroups? If possible, for $n$ not too big.
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For $n \ge 2$, $GL(n,\mathbb{Z})$ contains a subgroup isomorphic to the rank 2 free group $F_2$ (this is a very simple case of the Tits alternative). The commutator subgroup of $F_2$ is not finitely generated.
Lee Mosher
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What does this subgroup look like? Can you give explicit generators? – azimut Jul 07 '14 at 12:22
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2The upper and lower $2 \times 2$ unitriangular matrices with off-diagonal entries $2$ generate a free subgroup. – Derek Holt Jul 07 '14 at 12:43
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@DerekHolt: thanks. – azimut Jul 07 '14 at 15:09
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I want to use the same idea as Lee Mosher, based around the well-known free subgroup, but give a more explicit example (that is, the generators have a nicer form). In the free group $F(a, b)$ the set $\{a^kba^k:k\in\mathbb{Z}\}$ freely generates a free group on countably many generators (for a proof, see here). So, the matrices of the following form generate a non-finitely generated group of $GL(2,\mathbb{Z})$. $$ \left( \begin{array}{cc} 1&0\\ 2&1 \end{array} \right)^k \left( \begin{array}{cc} 1&2\\ 0&1 \end{array} \right) \left( \begin{array}{cc} 1&0\\ 2&1 \end{array} \right)^k $$ These matrices are easily seen to have the following form. $$ \left( \begin{array}{cc} 4k+1&2\\ 4k(2k+1)&4k+1 \end{array} \right) $$
As an aside, $\operatorname{GL}(n, \mathbb{Z})$ also contains finitely generated groups which are not finitely presentable. I cannot think of an obvious proof of this fact, and indeed my reasoning requires some serious results! The first result is about small cancellation groups. Small cancellation groups are groups with presentations where the relators do not interact very much with one another (any cancellation between the relators is "small"). It is a recent result, using some rather complicated machinery, that such groups are linear over $\mathbb{Z}$, so given a small cancellation group $G$ there exists some $n$ such that $G$ embeds into $\operatorname{GL}(n, \mathbb{Z})$. It is then a result of Rips that there exist small cancellation groups which contain finitely generated subgroups which are not finitely presentable: see Rips' paper Subgroups of small cancellation groups.
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2Not sure about $GL(3,\mathbb{Z})$, but if $n \ge 4$ then $GL(n,\mathbb{Z})$ contains a copy of $GL(2,\mathbb{Z}) \times GL(2,\mathbb{Z})$, and therefore a copy of $F_2 \times F_2$. The kernel of the "exponent sum" homomorphism $F_2 \times F_2 \to \mathbb{Z}$ is finitely generated but not finitely presented. – Lee Mosher Jul 08 '14 at 12:51
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That is a much prettier argument, thanks. However, I cannot see an easy reason why the subgroup you give is not finitely presentable - am I missing something? (Unless computing $H_2$ counts as "easy"?) – user1729 Jul 08 '14 at 13:06
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1The nicest proof I know is a baby example of Bestvina/Brady method. Letting $T$ be the Cayley tree for $F_2$, and $f : T \times T \to \mathbb{R}$ be an equivariant map with respect to the exponent sum homomorphism $F_2 \times F_2 \to \mathbb{Z}$, then you can see that $f^{-1}(0)$, which is quasi-isometric to a Cayley graph for the exponent sum kernel, is connected but has larger and larger "holes". Hence the kernel is finitely generated but not finitely presented. Still, though, this proof is really just computing $H_2$ (but in a powerful, conceptual, and generalizable way). – Lee Mosher Jul 08 '14 at 13:25
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Hmm, okay, thanks. I'm not awfully comfortable with these sorts of arguments, so I suppose this gives me something to chew on! – user1729 Jul 08 '14 at 13:32