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Let $G=GL_{n}(Z)$, the group of invertible matrices with entries in $Z$. Then show that

$G=${$A\in M_n(Z) : det(A)=1 \,\,\,or -1$}

Can you help me please?

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  • $\begingroup$ Hints: (1) $\det : M_n(\mathbb{Z})\to \mathbb{Z}$ is multiplicative and (2) the adjugate of a matrix with integer entries also has integer entries and the inverse of a matrix is the adjugate scaled by the reciprocal of the determinant. $\endgroup$ – Amitesh Datta Jan 18 '14 at 6:04
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Hint: The determinant of a product is the product of the determinants. And the determinant of an identity matrix is $1$.

For the other direction, one way is to use the formula for the inverse in terms of determinants.

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    $\begingroup$ Adding to your hint: Product of two integers is 1, then they are both +1 or both -1. $\endgroup$ – user44197 Jan 18 '14 at 6:04

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