Let $S:=\{A \in M_3(\mathbb Z) :A A^{\top}=I \}$ , then what is the cardinality of $S$? Let $S:=\{A \in M_3(\mathbb Z):A A^{\top}=I \}$, then what is the cardinality of $S$? 
My work: Since all the columns are orthonormal and also the rows are also orthonormal; so each column should have exactly one $+1$ or $-1$ and also each row should have exactly one $+1$ or $-1$; so in $1$st column there are $3$ possible locations for $+1$ or $-1$; so $2 \times 3$ choices; for the second column, there is only $2$ possible choice remaining for $+1$ or $-1$ because one row cannot have two non-zero entries; so $2\times 2$ is the number for 2nd column... thus a total of $2^33!$ choices. 
But this is almost direct counting; can we do it without counting? 
 A: Yes, this can be done more abstractly. Your $S$ is the hyperoctahedral group $O_3(\mathbb{Z})$. More generally the group $O_n(\mathbb{Z})$ can be viewed as the group of signed $n \times n$ permutation matrices, and it is a semidirect product $O_n(\mathbb{Z}) \cong (\{\pm1\})^n \rtimes S_n$ for we have a split short exact sequence
$$ 1 \to (\{\pm1\})^n  \to O_n(\mathbb{Z}) \to S_n \to 1, $$
where the monomorphism maps a sequence of elements of $\{\pm1\}$ to the corresponding diagonal matrix, the epimorphism forgets the signs, and a section lets us view a permutation in $S_n$ as a permutation matrix.
If we only care about cardinality, the algebraic nonsense above only says that every element in $O_n(\mathbb{Z})$ can be uniquely decomposed as a "sign vector" in $(\{\pm1\})^n$ and a permutation in $S_n$.
Thus the order of $O_n(\mathbb{Z})$ is the product of the orders of $(\{\pm1\})^n$ and $S_n$, i.e., $2^n n!$.
A: The fact that each row and each column contains exactly one non-zero coefficient tells you that any matrix in $S_n$ is of the form :
$$\begin{pmatrix}\lambda_1&&\\&\ddots&\\&&\lambda_n\end{pmatrix}M_{\sigma}$$
where $\sigma\in\mathfrak{S}_n$ and $M_{\sigma}$ is the associated permutation matrix and $\lambda_i\in \{\pm 1\}$. It is a straightforward verification that this leads to :
$$S_n\text{ is isomorphic to } \{\pm 1\}\wr \mathfrak{S}_n $$
That is $S_n$ can be expressed as a wreath product i.e. :
$$\{\pm 1\}^n\rtimes_{\text{ Action on coordinates}}\mathfrak{S}_n $$
In particular $|S_n|=2^nn!$.
