Problem from Galois' personal works Let $R =\mathbb Z /p\mathbb Z$, with $p $ prime, and consider the set $A $ of the permutations of $R $ of the form $$\rho_{a,b}:x \mapsto ax+b ,$$ with $a,b \in R$ and $a \ne 0$. Call $T $ the subset of $A $ of transformations with $a=1$.
I must show that every transitive solvable subgroup $L $ of $S_p $ is equivalent to a subgroup of $A $ (in the sense that $L $ is conjugate  to a subgroup of $A $ in $S_n $). The professor gave us some suggestions: basically that every transitive normal subgroup of $S_p $ is again a transitive group, and that if $G $ is a subgroup of $S_p $, then $T\unlhd G$ implies $G\le A$. So it seems that I should start like this: $G $ resolvable means that exist $\{G_i \}_{1\le i\le n}$ such that $1\unlhd G_1\unlhd\dots \unlhd G_n \unlhd G$ and every $G_i $ is maximal normal in $G_{i+1}$; since every $G_i $ must be transitive, $G_1$ must be the set of the $p$-cycles, that is $T $. This means that $G_2 $ is a subgroup of $A $, and from here I don't know how to proceed. Honestly I'm not very practical with the properties of normal and conjugate subgroups, so I'd like to have a hint to start from. Thank you in advance
 A: You need the following:
Claim: If $H$ is normal in $G \subset S_p$, and $G$ is transitive, then either $H$ is transitive or $H$ is trivial.
(You said something different "every transitive normal subgroup of $S_p$ is again a transitive group" which doesn't quite make sense, but you used the claim above, which is true so let's prove it)
Proof: Suppose that $X$ is an orbit under the action of $H$. Then $gX$ is also an orbit under the action of $H$, because $hgX = g(g^{-1}hg)X = gh'X = gX$. Since $G$ is transitive, it follows that all orbits of $H$ have the same size. Since these are acting on $p$ points, the orbits of $H$ must have size $1$ or size $p$. If the orbits have size $1$ then $H$ is trivial, otherwise $H$ is transitive.
Assume that $G$ is a transitive subgroup of $S_p$ which is also solvable. Exactly as you say, we may filter $G$ by subgroups $G_i$ each of which is normal in the next group, and whose successive quotients are simple and thus (by solvability) cyclic of prime order. By the claim above, $G_1$ is transitive, but it also has prime order, thus it must have order $p$. All order $p$ elements in $S_p$ are conjugate to $(1,2,\ldots,p)$, so after conjugating $G$ if necessary we may assume that $G_1$ is generated by this element.
Claim: $G_1$ is normal in $G_n$ for all $n$, and in particular is the unique $p$-Sylow subgroup of $G_n$. (At most one power of $p$ divides $|G_n|$ because only one power of $p$ divides $|S_p|$.)
Once you have this, you get that $G_1$ is normal in $G$, and so $G$ is contained in the normalizer of the $p$-Sylow. But the normalizer of $(1,2,\ldots,p)$ in $S_p$ is just $A$ (can you prove this? It follows from Sylow III) and so then (a conjugate of) $G$ is a subgroup of $A$, as desired. 
OK, so you asked for a hint, rather than a solution. The hint is to prove this by induction.
You can assume that $G_1$ is the unique $p$-Sylow of $G_{n-1}$, and that $G_1$ is normal in $G_{n-1}$, and that $G_n/G_{n-1}$ is generated by an element $\sigma \in G_n$ because the quotient is cyclic. So you really only need to show that $\sigma G_1 \sigma^{-1} = G_1$. Do you see why this must be so?
