# Galois's statement on a solvable primitive permutation group

The following statement is equivalent to the one Galois wrote in a paper submitted in 1830. Is this correct?

Let G be a finite solvable group acting faithfully and primitively on a set S. If a, b are distinct elements of S, then the point-wise stabilizer of {a, b} is 1.

I think this is false as the following examples show. Let p be a prime number. Let Z be the ring of rational integers. Let F = Z/pZ be the finite prime field. Let G = AGL(2, F) be the affine general linear group over F. G acts on F^2 faithfully and primitively. The point-stabilizer of {(0, 0), (0, 1)} is not 1. But G is solvable in case p = 2 or 3, as the order of G is (2^3)3 or (2^4)(3^3) respectively.

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S probably has prime order. – Jack Schmidt Apr 5 '12 at 23:23
Galois wrote clearly that the order of S was prime power p^m. Maybe if p > 3, his assertion is correct? – Makoto Kato Apr 6 '12 at 8:27
If a solvable group acts primitively on a finite set S, then S has prime power order p^k and G/C_G(S) is a subgroup of AGL(k,p). If S has prime order, then the stabilizer of any two points (as a tuple) is trivial, since AGL(1,p) has this property. This is often part of what is referred to as Galois's theorem. When k>2, there is always a solvable group that has a non-trivial 2 point stabilizer: AΓL(1,p^k), consisting of AGL(1,p^k) as well as the Frobenius automorphism of order k. – Jack Schmidt Apr 6 '12 at 12:27
Thanks. The followings are his original statements. 1) Pour qu'une equation de degre premier soit resoluble par radicaux, il faut et il suffit que deux quelconques de ses racines etant connues, les autres s'en deduisent rationnellement. 2) Pour qu'une equation primitive du degre m soit resoluble par radicaux, il faut que m = p^v, p etant un nombre premier. 3) A part les cas mentionnes ci-dessous, pour qu'une equation primitive du degre p^v soit resoluble par radicaux, il faut que deux quelconques de ses racines etant connues,les autres s'en deduisent rationnellement. – Makoto Kato Apr 6 '12 at 13:37
Hmm. Neither my French nor my Galois theory is all that great, but (3) really looks quite wrong, and I think this would be well known at the time, so probably this is our mistake not his. The symmetric group of degree 4 acting naturally is a counterexample (it is also known as AΓL(1,4)). For instance, the polynomial x^4-x-1 has roots a,b,c and the field Q[a] has degree 4, the field Q[a,b] has degree 12, and the field Q[a,b,c] is the splitting field of degree 24. Obviously the action is primitive. Maybe "primitive" means something else to Galois? – Jack Schmidt Apr 6 '12 at 15:32