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I can't solve this problem: Finite group in $GL(n,\mathbb{Q})$ is conjugate to finite group in $GL(n,\mathbb{Z})$. Could any one help me? Thanks a lot!

yang
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    That's not true once $n>1$; e.g. for $n=2$ the group of rotations by multiplies of $2\pi/m$ isn't conjugate to any group of $2 \times 2$ integer or even raitonal matrices except for $m=1,2,3,4,6$ because the trace is invariant and $\cos 2\pi/m$ is irrational. What is true is that a finite group of $n \times n$ matrices with rational entries is always conjugate to one with integer entries. – Noam D. Elkies Sep 14 '13 at 16:10
  • You are right! I'm very sorry that I made a mistake. – yang Sep 15 '13 at 01:01

1 Answers1

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As Noam D. Elkies has said, every finite subgroup of $GL(n,\mathbb{Q})$ is conjugated to a subgroup of $GL(n,\mathbb{Z})$. The proof is well known, see for example in the book of J.P. Serre on Lie Algebras and Lie Groups, Appendix $3$, Theorem $1$. A crucial lemma is, that for a finite subgroup $H$ of $GL(n,\mathbb{Q})$ there exists a lattice $M$ which sends $H$ onto itself.

Dietrich Burde
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  • I'm sorry that I don't understand his proof. Are there any other elegant and easy proof of this problem? – yang Sep 15 '13 at 01:03