Past exam of the options II course of the University of Oxford / 2010 / B / B2 / e

Table of contents
- Solution e

Solution

There are

p^{n}

possible translation vectors. An invertible matrix is specified by an ordered basis of

F_{p}^{n}

. Its first column can be any nonzero vector, giving

p^{n} - 1

choices. After choosing

i

independent columns, their span has

p^{i}

elements, so the next column has

p^{n} - p^{i}

choices. Therefore

∣ G ∣ = p^{n} ∣ GL_{n} (F_{p}) ∣ = p^{n} i = 0 \prod n - 1 (p^{n} - p^{i}) .

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