Exercise 14.3.12

Proof. Here

(a) Consider a complete system $\{a_1,\dots ,a_k\}$ of representatives of cosets of $V$ in ${𝔽}_p^n$, so that the cosets of $V$ are the disjoint cosets

Then ${𝔽}_p^n=\underset{i=1}{\overset{k}{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}}(a_i+V)$ is the disjoint union of parallel affine subspaces.

Let $g\in G\subset \mathrm{AGL}(n,{𝔽}_p)$. Recall (see the text following Corollary 14.3.18) that

thus $g={\gamma }_{A,w}$, for some $w\in {𝔽}_p^n$ and $A\in G_0$. For all $v\in V$

Since $A\in G_0$, $A\cdot V\subset V$, thus $g\cdot (a_i+v)\in (A\cdot a_i+w)+V.$ The cosets $R_i,1\le i\le k$, partition ${𝔽}_p^n$, thus there is some index $j$ such that $A\cdot a_i+w\in R_j=a_j+V$, so that $A\cdot a_i+w+V=a_j+V$, and

Moreover, the cosets $R_i=a_i+V$ have same cardinality, and $g$ is bijective, therefore $|a_j+V|=|a_i+V|=|g(a_i+V)|$. This proves

To conclude, there is a partition of ${𝔽}_p^n$ such that for every $1\le i\le k$, we have $g(R_i)=R_j$ for some $1\le j\le k$. Since $1<|V|<p^n$, then $k>1$ and $|R_1|=\cdots =|R_k|>1$. This proves that $G\subset S_{p^n}$ is imprimitive. (b) We have seen in Exercise 2 that the group homomorphism

induces an isomorphism

Consider the restriction ${\phi }_0$ of $\phi$, defined by

Since $G=\{{\gamma }_{A,w}\mid w\in {𝔽}_p^n,A\in G_0\}$, ${\phi }_0$ is surjective, and $\ker <!--\; FUNCTION\; APPLICATION\; -->{\phi }_0=\{{\gamma }_{I_n,w}\mid w\in {𝔽}_{p^n}\}\simeq {𝔽}_p^n$, thus, with the usual identification $\{{\gamma }_{I_n,w}\mid w\in {𝔽}_{p^n}\}={𝔽}_p^n$, ${𝔽}_p^n$ is a normal subgroup of $G$, and

Now we prove (b).

By Theorem 8.1.4, $G$ is solvable if and only if ${𝔽}_p^n$ and $G∕{𝔽}_p^n$ are normal. Since we know that ${𝔽}_p^n$ is a normal subgroup of $G$, we obtain the equivalence

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