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It is well known that $\mathbf{PGL}_2(\mathbb{Z})$ is finitely generated, and that $\mathbf{PGL}_2(\mathbb{Q})$ isn't. My question is: what is a fast, natural way to see these properties without explicit construction of generators?

For instance, do (some) generating properties of an integral domain (or division ring) $D$ carry over to $\mathbf{PGL}_2(D)$?

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    $\begingroup$ If $\mathrm{PGL}_2(\mathbf{Q})$ were f.g., it would equal $\mathrm{PGL}_2(\mathbf{Z}[1/n])$ for some $n\ge 1$. But this is not the case since the latter group misses the image of the matrix $e_{12}(1/p)$ for every prime $p$ not dividing $n$. This argument also works for an arbitrary infinite field (since no infinite field is a finitely generated ring), and, unlike any abelianization argument, works for $\mathrm{SL}_m$ as well. $\endgroup$
    – YCor
    Commented Jun 7, 2023 at 12:56
  • $\begingroup$ @YCor, good comment, but I am confused by your "unlike any abelianization argument"... Is there a typo? Or can you clarify? :) $\endgroup$
    – paul garrett
    Commented Jun 7, 2023 at 16:09
  • $\begingroup$ @paulgarrett I don't think there's a typo. The abelianization argument doesn't work for $\mathrm{SL}_m$. $\endgroup$
    – YCor
    Commented Jun 7, 2023 at 17:24
  • $\begingroup$ @YCor, ah, ok, good, thanks! Somehow I wasn't parsing it correctly. :) $\endgroup$
    – paul garrett
    Commented Jun 7, 2023 at 17:24
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    $\begingroup$ I'm willing to be that any fast, natural way to see that $\mathbf{PGL}_2(\mathbb{Z})$ is finitely generated without explicit construction of generators will just have an explicit construction of generators hidden in its proof. It's so simple to prove that fact about $\mathbf{PGL}_2(\mathbb{Z})$ that it has been generalized to the hilt, and all those generalizations are proved by, well, generalizing the construction for $\mathbf{PGL}_2(\mathbb{Z})$. $\endgroup$
    – Lee Mosher
    Commented Jun 13, 2023 at 0:59

2 Answers 2

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For $PGL_2(\mathbb{Z})$, you can use general properties of arithmetic groups due to Borel and Harish-Chandra, which will carry over to rings of $S$-integers of number fields. Alternatively, you can inspect the Smith Normal Form (aka elementary divisors) algorithm to obtain finite generation, which carries over to PIDs.

For $PGL_2(\mathbb{Q})$, note that every finitely generated subgroup is contained in $PGL_2(R)$ where $R$ is the subring generated by the entries of the generators, but $\mathbb{Q}$ is not finitely generated as a ring. This carries over to other rings that are not finitely generated. Alternatively, you can use the surjection given by the determinant $PGL_2(\mathbb{Q}) \to \mathbb{Q}^\times/(\mathbb{Q}^\times)^2$, where the target group is not finitely generated. This carries over to rings with infinitely generated group of units modulo squares.

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This is surely a matter of taste, but for me the fastest way to see that $PGL_2(\mathbb{Z})$ is finitely generated is to observe that it acts on the tree dual to the Farey graph with finite stabilizers and finite (i.e. compact) quotient. Therefore the Milnor–Schwarz Lemma implies that it is finitely generated.

Implicit in this, as Lee suggests, is a natural output of a choice of generators for any compact fundamental domain for the action.

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