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Is the following true ?

Every solvable transitive subgroup $G\subset\mathfrak{S}_p$ (the symmetric group on $p$ letters, where $p$ is a prime) contains a unique subgroup $C$ of order $p$ and is contained in the normaliser $N$ of $C$ in $\mathfrak{S}_p$. The quotient $G/C$ is cyclic of order dividing $p-1$. If $G$ is not cyclic, then it has exactly $p$ subgroups of index $p$.

I need such a result for an arithmetic application. A reference or a short argument will be appreciated.

Addendum. For those interested in the arithmetic application, see http://arxiv.org/abs/1005.2016

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A remark: this result is due to Galois himself, I believe. As an application: note that any subgroup of $AGL(1,p)$ that fixes two elements is trivial, and conversely, any transitive subgroup of $S_p$ with this property is contained in $AGL(1,p)$. Thus we get Galois's theorem: a prime degree irred. polynomial is solvable if and only if its splitting field is generated by any two roots. (This result was regarded as Galois's major contribution to the theory of equations for a decade or two after his death.) –  Emerton May 13 '10 at 5:42
    
Amazing! Great scholarship! I had guessed that such a result should be true from a reference to Artin's Galois Theory (Notre Dame notes), Chapter 3, Theorem 7, written by AN Milgram. He gives another nice corollary: a solvable irreducible prime-degree polymonial over a subfield of the reals has either (precisely) one real root or all its roots are real. –  Chandan Singh Dalawat May 13 '10 at 9:22
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Yes, that is a very nice corollary. For a reason I don't understand (because of the chronology: Galois is well before Kronecker), that result is traditionally attributed to Kronecker. A highly recommened reference for the history of these ideas is Hans Wussing's "The genesis of the abstract group concept". –  Emerton May 13 '10 at 14:36
    
I should add: Kronecker did publish a proof of that result; I just don't understand why it wasn't already seen at the time to be a consequence of Galois's more general result. –  Emerton May 13 '10 at 14:38
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Pour qu'une équation de degré premier soit résoluble par radicaux, il faut et il suffit que deux quelconques de ces racines étant connues, les autres s'en déduisent rationnellement (Évariste Galois, Bulletin de M. Férussac, XIII (avril 1830), p. 271). –  Chandan Singh Dalawat Jun 12 '10 at 5:57
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3 Answers 3

up vote 8 down vote accepted

A transitive subgroup $G$ of $S_p$ contains a Sylow $p$-subgroup $P$ having order $p$. If it has only the one, then $P$ is normal in $G$ and so $G$ lies in the normalizer $N$ of $P$ in $S_p$. This is the affine linear group $\mathrm{AGL}(1,p)$ which is soluble. Thus $G$ is soluble.

Otherwise $G$ has more than one Sylow $p$-subgroup. By Sylow's theorems, these $p$-subgroups are conjugate in $G$. If $H$ is a nontrivial normal subgroup of $G$ then $H$ must be transitive since $G$ is primitive (the orbits of $H$ form a partition invariant under the action of $G$). So $H$ contains a Sylow $p$-subgroup $P$ of $G$. So $H$ contains all the Sylow $p$-subgroups of $G$ (as they are conjugate under $G$). Therefore $G$ cannot be soluble, as by repeatedly taking nontrivial normal subgroups we always get groups with more than one Sylow $p$-subgroup.

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Many thanks for a convincing argument. –  Chandan Singh Dalawat May 11 '10 at 4:35
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This is exercise 7.2.12 of Robinson's Course in the Theory of Groups, page 195 in the first edition.

A transitive subgroup of prime degree is primitive, and primitive solvable groups have a regular normal subgroup that is complemented by a unique conjugacy class of maximal subgroups. In particular, the Sylow p-subgroup C of order p is that regular normal subgroup, and the complement (being a permutation group) acts faithfully on it. In other words, the centralizer of the subgroup C is C itself. Hence G/C is a subgroup of Aut(C), so cyclic of order dividing p-1.

Since G is solvable it has a Sylow p-complement M, and by Hall's 1928 theorem, the number of such Sylow p-complements is a divisor of p. If it is 1, then M is normal, so M centralizes C, so M=1, and G=C is cyclic.

This is summarized by saying that G is a subgroup of AGL(1,p) containing the translation subgroup. More generally every primitive solvable group is a subgroup of AGL(n,p) where p^n is the degree of the permutation action (but AGL(n,p) is no longer solvable itself).

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Many thanks for the argument and the reference. –  Chandan Singh Dalawat May 11 '10 at 4:35
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Happened to come across the following Satz in Huppert, Endliche Gruppen I, [S. 163].

enter image description here

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