Exercise 14.3.21

Proof. We write $q=p^m$. $\text{∙}$ Let $\mathcal{}B=(e_1,\dots ,e_m)$ be a base of ${𝔽}_q$ over ${𝔽}_p$. As ${𝔽}_p$-vector spaces, ${𝔽}_q^n$ and ${𝔽}_p^{\mathit{nm}}$ are isomorphic, where an isomorphism $\phi$ is given by

where ${\alpha }_i={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^m{x}_j^i{e}_j,i=1,\dots ,n.$ (this isomorphism depends of the choice of the base $\mathcal{}B$). So we can write

$\text{∙}$ ${𝔽}_{p^m}^n={𝔽}_q^n$ is identified with $\{{\gamma }_{I_n,v}\mid v\in {𝔽}_q^n\}\subset \mathrm{AGL}(n,{𝔽}_q)$, via the isomorphism

This shows that

$\text{∙}$ If $f={\gamma }_{A,v}$ is any element of $\mathrm{AGL}(n,{𝔽}_q)$, then $f={\gamma }_{A,e,v}\in \mathrm{A\Gamma L}(n,{𝔽}_q)$ (where $e$ is the identity is $S_q$). Therefore

$\text{∙}$ By Exercise 3(c), we know that elements of $\mathrm{A\Gamma L}(n,{𝔽}_q)$ give maps ${𝔽}_q^n={𝔽}_p^{\mathit{nm}}\to {𝔽}_q^n={𝔽}_p^{\mathit{nm}}$ that are affine linear over ${𝔽}_p$, so that these elements are in $\mathrm{AGL}(\mathit{nm},{𝔽}_p)$.

$\text{∙}$ The elements of $\mathrm{AGL}(\mathit{nm},{𝔽}_p)$ are bijective maps from ${𝔽}_p^{\mathit{nm}}$ to ${𝔽}_p^{\mathit{nm}}$, thus $\mathrm{AGL}(\mathit{nm},{𝔽}_p)\subset S({𝔽}_p^{\mathit{nm}}).$

Via an arbitrary numbering of ${𝔽}_p^{\mathit{nm}}$, say $\gamma :\{1,\dots ,p^{\mathit{nm}}\}\to {𝔽}_p^{\mathit{nm}}$, we obtain the isomorphism $\psi :S({𝔽}_p^{\mathit{nm}})\to S_{\mathit{nm}}$ defined by $\psi (\sigma )={\gamma }^{-1}\circ \sigma \circ \gamma$, which allows us to identify $S({𝔽}_p^{\mathit{nm}})\simeq S_{\mathit{nm}}$.

To conclude, using several improper non-canonical identifications, we write

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