Exercise 14.3.17

Proof. We have seen in Exercise 2 that $\mathrm{AGL}(n,{𝔽}_p)∕{𝔽}_p^n\simeq \mathrm{GL}(n,{𝔽}_p)$, thus

To construct a matrix $A\in \mathrm{GL}(n,{𝔽}_p)$, we must choose the successive columns $c_i$ of the matrix $A$, so that $(c_1,\dots ,c_n)$ is a base of ${𝔽}_p^n$. This is equivalent to take $c_1\ne 0$, then $c_2\notin ⟨c_1⟩$, then $c_3\notin ⟨c_1,c_2⟩$, $\text{…}$, up to $c_n\notin ⟨c_1,\dots ,c_{n-1}⟩$. Since $|⟨c_1,\dots ,c_k⟩|=p^k$, when $c_1,\dots ,c_k$ are linearly independent, we obtain $|\mathrm{GL}(n,{𝔽}_p)|=(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})$, thus

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