Finite subgroups of $O_n(\mathbb{Z})$ versus $O_n(\mathbb{Q})$ Are there any cases of finite subgroups of $O_n(\mathbb{Q})$ not contained in not isomorphic to any subgroup of $O_n(\mathbb{Z})$?
 A: Let $n=m^2$ be a square. Then the vector $(1,\dots,1)\in\mathbf{Q}^n$ has norm $m$, as does $(0,\dots 0,m)$. Hence by Witt's theorem there exists an element $u$ of $\mathrm{O}_n(\mathbf{Q})$ mapping $(0,\dots,0,m)$ to $(1,\dots,1)$. Hence $u$ maps orthogonals to orthogonals: it maps the hyperplane $\mathbf{Q}^{n-1}\times\{0\}$ of equation $x_n=0$ onto the hyperplane of equation $\sum x_i=0$. Let $S_n$ be the subgroup of $\mathrm{O}_n(\mathbf{Q})$ permutating coordinates. Then $S_n$ fixes $(1,\dots,1)$ and acts faithfully on its orthogonal. Therefore $u^{-1}S_nu$ acts faithfully on the orthogonal $\mathbf{Q}^{n-1}$.
Hence, for every square $n$, the group $\mathrm{O}_{n-1}(\mathbf{Q})$ contains a copy of the symmetric group $S_n$. But for every $n\ge 5$ there is no nontrivial homomorphism $S_n$ to the Weyl group $S_{n-1}\ltimes C_2^{n-1}$ of type $B_{n-1}$, which is isomorphic to $\mathrm{O}_{n-1}(\mathbf{Z})$.
This proves that for every square $n\ge 9$, there is a finite subgroup of $\mathrm{O}_{n-1}(\mathbf{Q})$, isomorphic to $S_n$, not embedding into $\mathrm{O}_{n-1}(\mathbf{Z})$.
A: A theorem which you might be looking for is that, if $G$ is any finite subgroup of $O_n(\mathbb{Q})$, then there is a lattice $\Lambda$ in $\mathbb{Q}^n$ which is preserved by $G$. However, that lattice does not have to be $\mathbb{Z}^n$.
Proof: Let $L$ be any lattice in $\mathbb{Q}^n$. Take $\Lambda = \sum_{g \in G} g L$.
A: If it's the same $n$ then yes this can happen.  For example, the lattice $D_4$ (consisting of all integer vectors in ${\bf Z}^4$ with even sum) has more isometries than ${\bf Z}^4$. If we allow different $n$, then no, because every finite group $G$ is contained in the  group of permutations of $G$, which in turn is contained in $O_{|G|}({\bf Z})$.
