If I consider an element $A \in SL(2,\mathbb{Z})$, then I have that $A^{-1}\in SL(2,\mathbb{Z})$. I can see this because the inverse of $A$ is obtained by movin the coefficient of the metrix or changing their sign.
Does the same hold for an element of $SL(d,\mathbb{Z})$. I cannot convince myself of that. Could you explain this to me or give ma counterexample?
I cannot convince myself of these even thinking of lattice automorphism of $\mathbb{Z}^d$
The cofactors of $A$ are obtained by performing ring operations on the entries of $A$. The inverse of $A$ is then the matrix of cofactors of $A$ divided by the determinant of $A$. Since the determinant of $A$ is $\pm 1$, all of the entries of $A^{-1}$ will lie in $\mathbb{Z}$ and the determinant will be the same as that of $A$.
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