I asked the following question (https://math.stackexchange.com/questions/1487961/reference-for-every-finite-subgroup-of-operatornamegl-n-mathbbq-is-con) on math.stackexchange.com and received no answers, so I thought I would ask it here. I've asked several people in my department who were all stumped by the question.

The question is: why is every finite subgroup of $\operatorname{GL}_n(\mathbb{Q})$ conjugate to a finite subgroup of $\operatorname{GL}_n(\mathbb{Z})$?

Note that at least for $n=2$ the question of isomorphism is much easier, since one can (with some effort) work out exactly which finite groups can be subgroups of $\operatorname{GL}_2(\mathbb{Q})$. Further, there are isomorphic finite subgroups of $\operatorname{GL}_2(\mathbb{Q})$ that are not conjugate to each other. For example, the group generated by $-I_{2 \times 2}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ are both isomorphic to $C_2$, but they cannot be conjugate to each other because the eigenvalues of the two generators are different.

If there is a relatively simple proof, that would be ideal, but a reference with a potentially long proof is fine as well.

Thanks for any assistance.

  • $\begingroup$ Can I ask how do you know it's true? $\endgroup$ – Todd Trimble Oct 20 '15 at 2:05
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    $\begingroup$ @ToddTrimble this was posted in the original question, someone casually mentioned it without justification here: math.stackexchange.com/questions/58476/… . Nobody seemed to contest that claim, so I assumed it was some well-known result in the subject. I've asked several faculty members in my department, none of them knows why it's true, but all of them believe it $\endgroup$ – Stanley Yao Xiao Oct 20 '15 at 2:07
  • $\begingroup$ Frankly, had you tagged it correctly on math.se and waited a bit you'd gotten your answer there I am pretty sure. $\endgroup$ – user9072 Oct 20 '15 at 16:36

This argument is fairly standard, but it is quicker to repeat it than to find a reference: Let $G$ be a finite subgroup of $GL_n(\mathbb{Q})$. Set $\Lambda = \sum_{g \in G} g \cdot \mathbb{Z}^n \subset \mathbb{Q}^n$. Then $\Lambda$ is a finitely generated torsion free abelian group, hence isomorphic to $\mathbb{Z}^r$ for some $r$. Since $\mathbb{Z}^n \subseteq \Lambda \subset \mathbb{Q}^n$, we have $r=n$. Clearly, $g \Lambda = \Lambda$ for all $g \in G$.

Let $h \in GL_n(\mathbb{Q})$ take the standard basis to a basis of $\Lambda$, so $h \mathbb{Z}^n = \Lambda$ and $\mathbb{Z}^n = h^{-1} \Lambda$. Then $h^{-1} g h$ takes $h^{-1} \Lambda = \mathbb{Z}^n$ to itself for all $g \in G$, so $h^{-1} G h \subset GL_n(\mathbb{Z})$.


David Speyer has answered the question, but let me add some background. The general result is that if $R$ is a principal ideal domain with field of fractions $K$, and $G$ is a finite group, then every finite dimensional representation of $G$ over $K$ may be realised over $R$ ( ie, is equivalent to a representation over $R$). The general proof is essentially the one that David gives.

As well as the integral case considered in the question, the general result was important for the development by Richard Brauer of modular representation theory: the idea of "reduction (mod $p$)" of a complex representation relies on it: the representation of the finite group $G$ is first realised over a suitable finite extension $K$ of $\mathbb{Q}$, and then $K$ may be viewed as the field of fractions of a localisation $R$ at a prime ideal $\pi$ containing $p$ of the ring of algebraic integers in $K$. Then since $R$ is a PID, the $K$-representation may then be realized over $R$. Then the given representation may be reduced (mod $\pi$), yielding representation of $G$ over the finite field $R/\pi$.

In general, it is not a straightforward issue to decide whether a representation of a finite group $G$ over a number field $K$ may be realised over the ring of integers of $K$. Some of the issues are well illustrated in the article "Three letters to Walter Feit" by J-P. Serre (which is visible online), which considers special cases of this question.


I think this should work: let $S$ be the set of primes occurring in the denominators of the entries of the elements of $G$, and let $\mathbb{Z}[S^{-1}]$ be the subring of $\mathbb{Q}$ generated by adjoining $1/p$ for $p\in S$.

Let $\mathbb{Q}_S = \prod_{p\in S} \mathbb{Q}_p$ and let $\mathbb{Z}_S = \prod_{p\in S} \mathbb{Z}_p$. We'll show that we can conjugate $G$ into $GL_n(\mathbb{Z}_S)$ by an element of $x$ $GL_n(\mathbb{Z}[S^{-1}])$. This is enough, since it shows then the $p$-adic valuation of all entries of elements of $x^{-1}Gx$ is nonnegative, and of course we still have $x^{-1}Gx\in GL_n(\mathbb{Q})$.

To this end, because $G$ is finite, it compact, so given a $p$ there is some $x_p\in GL_n(\mathbb{Q}_p)$ such that $x_p^{-1}Gx_p \leq GL_n(\mathbb{Z}_p)$. In fact, the set of such $x_p$ is open in $GL_n(\mathbb{Q}_p)$ since $G$ is finite and the map $x_p \mapsto x_p^{-1}gx_p$ is continuous. Because $GL_n(\mathbb{Z}[S^{-1}])$ is dense in $GL_n(\mathbb{Q}_S)$, there is in fact an $x\in GL_n(\mathbb{Z}[S^{-1}])$ that conjugates $G$ into $GL_n(\mathbb{Z}_S)$, completing the proof.

  • $\begingroup$ I think the fact that every compact subgroup of $\operatorname{GL}_n(\mathbb Q_p)$ is conjugate into $\operatorname{GL}_n(\mathbb Z_p)$ is not necessarily easier than the proof of the desired fact given by @DavidSpeyer above. $\endgroup$ – LSpice Oct 21 '15 at 19:57
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    $\begingroup$ Yeah, after I saw @David's proof I agree. $\endgroup$ – John Binder Oct 21 '15 at 21:19

This is not exact answer to your question, but long back, I had seen this result in an article in Monthly. I hope it will be useful.


  • $\begingroup$ Yes, a professor in my department pointed that paper out to me some time after David Speyer's anwer. Thanks for the link $\endgroup$ – Stanley Yao Xiao Jan 5 '16 at 3:22

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