Is $GL(2,\mathbb Z)$ a group? Is the set of $2\times 2$ invertible matrices with integer entries a group under matrix multiplication? I believe not, because inverses for elements in this set may not be in the set (ie, may not have integer entries). Is this correct?
 A: For any ring $A$, commutative, with $1$,
$GL(2,A)$ is the set of $2\times 2$ matrices with elements in $A$ with determinant an invertible element of $A$. For $A= \mathbb{Z}$ we get $GL(2,\mathbb{Z})$ the set of $2\times 2$ matrices with integral elements and determinant $\pm 1$. For $A=F$ a field we get us $GL(2,F)$ the $2\times 2$ matrices with elements in $F$ and determinant $\ne 0$. 
So yes,$\ GL(2,A)$ is  a group.
A: The general linear group $GL(2,\mathbb{Z})$ of order 2 over the integers is a proper subset of the $2\times 2$ integer matrices that are invertible as real or rational matrices.  
A $2\times 2$ matrix with integer entries may be invertible (nonzero determinant) but the inverse will have integer entries only if the determinant is $\pm 1$.  The larger set of matrices do form a cancellative monoid.
Added:  Using the fact that determinant of a product is the product of determinants, one can easily show that for an integer $n\times n$ matrix to have an inverse that is also an integer $n\times n$ matrix the determinant must be an integer that is a unit, i.e. $\pm 1$.  That is, if $BC = I$, then $|B| |C| = |I| = 1$, and if $C$ has integer entries, $|C|$ must be an integer.
This easily generalizes to the case of entries restricted to a commutative ring $R$ with identity, in that the determinants must be units in the ring if the matrix inverse is to have entries in the ring $R$.
