# Galois and solvable primitive permutation groups

I posted on this subject recently, but there was a misunderstanding on my side. Since my French is not very good, I misread the Galois's original paper. So let me explain my question again.

Let $f(X)$ be an irreducible polynomial over a field $K$ of characteristic $0$.
Let $L$ be a splitting field of $f(X)$ over $K$.
Let $G$ be the Galois group of $L/K$.
Let $S$ be the set of all roots of $f(X)$ in $L$.
Suppose $G$ is solvable and it acts primitively on $S$.

It is well-known that the degree of $f(X)$ is a prime power $p^k$.
Let $a, b$ be distinct elements of $S$.
Galois wrote that the pointwise stabilizer of $\{a, b\}$ is $1$ except the following cases.

1. $p^k = 9, 25$

2. $p^k = 4$

3. $k > 1$ and let $(p^k - 1)/(p - 1)$ be $M$. There is a prime power divisor $q^r$ of $M$ such that $(M/q^r)k = p \pmod q^r$

He wrote something about the pointwise stabilizer of $\{a, b\}$ in this case, but I cannot understand it fully. He wrote (I changed a bit of notations):

il faudra toujours que deux des racines etant connues, les autres s'en deduisent, du moins au moyen d'un nombre de radicaux, du degre $p$, egal au nombre des diviseurs $q^r$ de $M$ qui sont tels que $(M/q^r)k = p \pmod q^r$, $q$ premier.

It seems to me that he meant the pointwise stabilizer of $\{a, b\}$ is of order $p^s$ and $s$ is the number of prime power divisors $q^r$ which satisfies the above condition.

Are these correct? Could someone please explain these to me?

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It should always be the case that, should two of the roots be known, the others can be deduced, with the help of a number of radicals of degree $p$, equal to the number of divisors $q^r$ of $M$ which are such that $(M/q^r)k \equiv p$ mod $q^r$, $q$ prime.