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Let $G$ be a solvable group and $X$ is a $G$-set which is transitive and faithful, with $|X|=p$ is prime. The aim is to prove

There is an injective homomorphism $i:G\to Aff(\mathbb{F}_p)$ and bijective $j:X\to\mathbb{F}_p$ such that $j(g\cdot x)=i(g)\cdot j(x)$.

(Intuitively, we want to give the set $X$ a group structure -- which then must be isomorphic to $\mathbb{F}_p$ -- such that the $G$-action is just an affine map, i.e., of the form $x\mapsto ax+b$.)

Let $G=G_n\triangleright\cdots\triangleright G_0=\{1\}$ be the composition series of $G$ (so $G_{i+1}/G_i$ is cyclic of prime order). I've proved that each $G_k$ action on $X$ is transitive and faithful. So we can proceed by induction on $n$, assuming the analogous result is true for $G_0,\ldots,G_{n-1}$.

For $n=1$, in fact the injection $i_1:G_1\to Aff(\mathbb{F}_p)$ can be chosen so that $i_1(g)$ is an affine map of the form $x\mapsto x+b$. Then I also proved $G_1\triangleleft G_n$. Let $x_0=j^{-1}(0)$. Consider the subgroup $H=Stab(x_0)$ of $G_n$. This is where I get stucked:

Prove that there is an injective homomorphism $\tilde{i}:H\to Aff(\mathbb{F}_p)$ such that $j(g\cdot x)=\tilde{i}(g)\cdot j(x)$ for all $g\in H,x\in X$, and, furthermore, $\tilde{i}(g)$ is an affine map of the form $x\mapsto ax$ for all $g\in H$.

If this is proved, I think I can finish the rest of the proof.

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It $|G|=p$ is a prime number, then $G$ is simply a cyclic group. Maybe you mean something else, for instance $|X|=p$? –  Dan Shved Nov 7 '12 at 7:42
    
Right, I meant $|X|=p$. Thanks. –  mitt12598 Nov 7 '12 at 8:05
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1 Answer 1

Up to now your results are (I leave out a lot of $j$ by actually identifying): $N\lhd G$ with $N\cong\mathbb F_p$ operates on $\mathbb F_p$, where $N=\langle n\rangle$ operates by addition, i.e. $n\cdot x=x+1$.

We need to show that $g\in G$ with $g\cdot 0 =0$ operates as multiplication with $r$ where $gng^{-1}=n^r$. For $x\in \mathbb F_p$ we have $x=n^k\cdot 0$ for some $k$ and then $$g\cdot x=gn^k\cdot 0 = (gng^{-1})^kg\cdot 0 = (gng^{-1})^k\cdot 0=n^{rk}\cdot0=\underbrace{x+\cdots +x}_r=rx.$$

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