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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?

Joe
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bsm
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    While your reasoning is correct, the notation $GL(2,\Bbb Z)$ usually denotes the invertible matrices over $\Bbb Z$. That is, their inverses also have integral entries. – Quang Hoang Sep 08 '14 at 03:09
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    It depends on what you mean by "invertible"! – Qiaochu Yuan Sep 08 '14 at 04:26
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    As others have commented, this is really a matter of interpretation of conventions. The group of $2 \times 2$ invertible rational matrices which happen to have integer entries do NOT form a group ( eg $\left( \begin{array}{clcr}2 & 0\0&1 \end{array} \right)$ is an invertible rational matrix, but its inverse does not have integer entries. The usual convention is that for a commutative ring $R$, ${\rm GL}(n,R)$ is the set of $n \times n$ matrices with entries in $R$ with determinant is a unit in $R$. This is always a group. When $R = \mathbb{Z}$, this means det $\pm 1$ not det $\neq 0.$ – Geoff Robinson Sep 08 '14 at 08:20

2 Answers2

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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.

orangeskid
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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 $R$.

hardmath
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  • Thanks, @TedShifrin , I did see that Comment and upvoted it. – hardmath Sep 08 '14 at 04:19
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    And, again, it perhaps bears repeating and emphasizing that the notation $GL(2,R)$ for a commutative ring $R$ is most often taken to refer to matrices with entries in $R$ and determinant in $R^\times$, ... thus pointedly skirting the large possibility that $R^\times$ is much smaller than $R-{0}$. – paul garrett Sep 08 '14 at 12:34