Let $G$ be the group of all $n\times n$ matrices with entries of the matrices from the field $\mathbb Z_p$ , $p$ prime, such that determinant of every matrix is not $[0]$ , w.r.t. matrix multiplication ; then what is the order of $G$ ?
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Ok, how many ordered basis are there for $;\left(\Bbb Z_p\right)^n;$ over $;\Bbb Z_p;$ – Timbuc Nov 29 '14 at 12:46
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possible duplicate of Number of bases of an n-dimensional vector space over q-element field. and http://math.stackexchange.com/questions/859520 – Martin Brandenburg Nov 29 '14 at 13:08
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Further hint:
For any matrix in $\;G\;$ , we look at its columns. The first one cannot be zero, so there are $\;p^n-1\;$ possible choices for it.
Now, the second column must be linearly independent of the first one, so we must avoid all its $\;p\;$ scalar multiples (pay attention to the fact that this already takes care of a possible zeros column), so there are $\;p^n-p=p\left(p^{n-1}-1\right)\;$ possible choices for this second column.
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Timbuc
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It is easier to count $\text{SL}_n\left(\mathbb{Z}_p\right)$ and consider its cosets with non-zero determinant.
John McGee
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