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Let $O_n(\mathbb Z)$ be a group of orthogonal matrices B st. $B*B^T=I$ with entries $b_{ij} \in \mathbb Z$. How do I show that $O_n(\mathbb Z)$ is a finite group and find its order. 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 $I \in O_n(\mathbb Z)$.

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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.

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  • $\begingroup$ How is this going to help me show symmetric group is subgroup of $O_n(Z)$ ? @Michael $\endgroup$ – abcdef Oct 6 '15 at 12:07
  • $\begingroup$ Do you know of matrices with $0$s and $1$s which permute numbers? $\endgroup$ – Empy2 Oct 6 '15 at 13:04
  • $\begingroup$ Not really @Michael $\endgroup$ – abcdef Oct 6 '15 at 15:02
  • $\begingroup$ Look up permutation matrix $\endgroup$ – Empy2 Oct 6 '15 at 16:13
  • $\begingroup$ I see that it fulfills requirements for being a group how do i show it is a subgroup O(Z)? @Michael And here I am speaking for Symmetric groups $S_n$ not permutation matrix, $\endgroup$ – abcdef Oct 7 '15 at 20:01
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Let, $A$ an orthogonal $n \times n$ matrix with integer entries. First to all, we know that $\det(A)=\pm 1$. That's mean $A$ must be invertible matrix.

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

When, $A$ has only positive coordinates. $A$ is a permutation matrix, is not difficult to see that thoose matrices are isomorfic to $S_n$.

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