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 $I \in O_n(\mathbb Z)$.
Best Answer
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.