Are there any cases of finite subgroups of $O\_n(\mathbb{Q})$ ~~not contained in~~ *not isomorphic to any subgroup of* $O\_n(\mathbb{Z})$?



Finite subgroups of $O_n(\mathbb{Z})$ versus $O_n(\mathbb{Q})$
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 $\mathrm{O}\_{n-1}(\mathbf{Z})$.


