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It is known that any finite subgroup of $GL(n, \mathbb{Z})$ is isomorphic to a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for any odd prime $p$ (see here). I am wondering if there is a converse to to this: Does $G$ being isomorphic to a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for all odd primes $p$ imply it is isomorphic to a subgroup of $GL(n, \mathbb{Z})$?

If this does hold, I'd be curious if it could be strengthened (perhaps being a subgroup for any infinite set of primes, or all but finitely many primes, would work), and if it doesn't, I'd be curious if any weaker form could be salvaged.

YCor
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Beren Gunsolus
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  • In the original implication, the finite subgroups of the $GL(n,\Bbb F_p)$s are all connected in that they are images of a subgroup of $GL(n,\Bbb Z)$. In your converse, there seems to be no connection between these subgroups, which makes the question a bit "unnatural" in my opinion. (Also, why is the word images in my comment not being italicized?) – anon Jan 09 '21 at 07:53
  • I agree that it's less natural -- nevertheless, I'm personally curious to know the answer! – diracdeltafunk Jan 09 '21 at 07:54
  • @runway44 i don't find this unnatural at all -- it is very much in the spirit of e.g. the hasse principle for quadratic forms: here's a bunch of local obstructions to a phenomenon, if it is unobstucted locally everywhere is this enough to prove it is unobstructed globally? – hunter Jan 09 '21 at 07:55
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    @hunter I get that the converse is in that vein, but in your example all the quadratic forms at the local places are connected in that they come from a quadratic form over a number field. I don't think the subgroups of $GL(n,\Bbb F_p)$s being merely isomorphic makes them comparably connected to each other. – anon Jan 09 '21 at 07:58
  • We have a family ${ \phi_p }$ of representations of the group in characteristic $p$ on a two-dimensional vector space and ask if we can recover a $\phi$ reducing to them. It is a beautiful question! – hunter Jan 09 '21 at 07:59

2 Answers2

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No, the quaternion group $Q_8$ is a subgroup of ${\rm GL}(2,p)$ for all odd primes $p$, but not of ${\rm GL}(2,{\mathbb Z})$.

Derek Holt
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    For context: let $K$ be a field with $0\neq 2$. Then, by an elementary verification, $\mathrm{GL}(2,K)$ contains a copy of $Q_8$ iff $-1$ is a sum of two squares in $K$. This applies to every finite field of odd cardinal (and hence every field of odd positive characteristic), but not to $\mathbf{Q}$. – YCor Jan 09 '21 at 13:28
  • I cannot answer your comment. In French and maths, the same word is used for the two different meanings. – Claude Leibovici Jan 13 '21 at 11:52
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Here's one thing we can say. Suppose a finite group $G$ embeds into $GL_n(\mathbb{F}_p)$ for all $p$; actually we can ask for the much weaker condition that $G$ embeds into $GL_n(\prod \mathbb{F}_p)$, and also weaken $G$ to be finitely generated. Then $G$ embeds into $GL_n(K)$ where $K$ is the quotient of $\prod \mathbb{F}_p$ by any of its maximal ideals containing the ideal of elements with finite support; these correspond to ultraproducts $\prod \mathbb{F}_p/U$, and importantly are fields of characteristic $0$.

Since $G$ is finitely generated, this embedding actually takes values in the finitely generated subring $R$ of $K$ generated by the matrix entries of a finite set of generators of $G$. So $G$ embeds into $GL_n(R)$ where $R$ is a finitely generated integral domain of characteristic $0$. This sort of manipulation is a key step in the proof of Malcev's theorem that a finitely generated linear group is residually finite.

Qiaochu Yuan
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  • Is there any single such integral domain that can handle any group, preferably as simple as possible? – Beren Gunsolus Jan 10 '21 at 04:36
  • @B Gunsolus: do you mean "any finite group"? Asking for any group is much too strong. Even for finite groups the answer is no. You can show that any finite group $G$ which embeds into $GL_n(R)$ for $R$ an integral domain of characteristic $0$ embeds into $GL_n(\mathbb{C})$ so we can work over $\mathbb{C}$ WLOG. Now the Jordan-Schur theorem implies that a finite subgroup of $GL_n(\mathbb{C})$ has a normal abelian subgroup of some index $f(n)$ which is a function of $n$ only: https://en.wikipedia.org/wiki/Jordan%E2%80%93Schur_theorem – Qiaochu Yuan Jan 10 '21 at 04:46
  • Various other arguments are possible. For example it's known that the smallest faithful representation of the symmetric group $S_n$ has dimension $n-1$ for $n \ge 7$, so $S_{n+2}$ doesn't embed into $GL_n(\mathbb{C})$ (and hence into $GL_n(R)$ for $R$ an integral domain of characteristic $0$) for $n \ge 5$. In characteristic $p$ we can run a similar but somewhat simpler argument which you can find here: https://math.stackexchange.com/a/3845230/232 – Qiaochu Yuan Jan 10 '21 at 04:51
  • Yes, I meant finite groups. Thank you! – Beren Gunsolus Jan 10 '21 at 05:05