Order of the group of integer orthogonal matrices

Question

Let $O_n(\mathbb Z)$ be the group of orthogonal matrices (matrices $B$ s.t. $BB^T=I$) with entries in $\mathbb Z$.
1) How do I show that $O_n(\mathbb Z)$ is a finite group and find its order?
2) I need to show also that symmetric group $S_n$ is a subgroup of $O_n(\mathbb Z)$.

So it needs to satisfy associativity/identity/inverse.

It is easy to see that every orthogonal matrix $A \in O(\mathbb Z)$ has an inverse, namely $A^T$. Moreover, the product of two orthogonal matrices is orthogonal since $(AB)^T = B^T A^T$. If $A, B \in O_n(\mathbb Z)$ then $(AB)^T(AB) = B^T A^T AB = BIB^T = BB^T = I$, hence $O_n(\mathbb Z)$ is closed under multiplication, since $I \in O_n(\mathbb Z)$.

Answer 1

HINT: Every column vector has length $1$, and all $b_{ij}$ are integers, so exactly one of $b_{i1}$ is $\pm1$ and all the others are zero.

Answer 2

Let $A$ be an orthogonal $n \times n$ matrix with integer entries. First of all, we know that $\det(A)=\pm 1$. This means that $A$ must be an invertible matrix.

Secondly, let $A_1,A_2,\dots,A_n \in \mathbb{Z}^n$ be the columns vectors of $A$. $A$ is orthogonal so $||A_i||=1$ for $i=1,2,...,n$. Then $a_{i1}^2+\cdots+a_{in}^2=1$, $A_i \in \mathbb{Z}^n$, hence $A_i$ is a canonical vector for all $i$. The vectors $A_i$ must be an independent set. Finally those matrices have the following form: $$ A=\Bigg[ \pm e_{\sigma(1)} \pm e_{\sigma(2)}\dots\pm e_{\sigma(n)} \Bigg], $$ where $\sigma(i)$ is a permutation of $I_n$.

When $A$ has only positive coordinates, $A$ is a permutation matrix, and it is not difficult to see that those matrices are isomorphic to $S_n$.