Solvable transitive groups of prime degree - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:39:51Z http://mathoverflow.net/feeds/question/24081 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24081/solvable-transitive-groups-of-prime-degree Solvable transitive groups of prime degree Chandan Singh Dalawat 2010-05-10T11:26:55Z 2011-10-14T04:01:43Z <p>Is the following true ?</p> <blockquote> <p>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 $p$.</p> </blockquote> <p>I need such a result for an arithmetic application. A reference or a short argument will be appreciated.</p> <p><strong>Addendum.</strong> For those interested in the arithmetic application, see <a href="http://arxiv.org/abs/1005.2016" rel="nofollow">http://arxiv.org/abs/1005.2016</a></p> http://mathoverflow.net/questions/24081/solvable-transitive-groups-of-prime-degree/24089#24089 Answer by Jack Schmidt for Solvable transitive groups of prime degree Jack Schmidt 2010-05-10T12:04:47Z 2010-05-10T12:04:47Z <p>This is exercise 7.2.12 of Robinson's Course in the Theory of Groups, page 195 in the first edition.</p> <p>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.</p> <p>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.</p> <p>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).</p> http://mathoverflow.net/questions/24081/solvable-transitive-groups-of-prime-degree/24091#24091 Answer by Robin Chapman for Solvable transitive groups of prime degree Robin Chapman 2010-05-10T12:33:26Z 2010-05-10T12:33:26Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/24081/solvable-transitive-groups-of-prime-degree/58906#58906 Answer by Chandan Singh Dalawat for Solvable transitive groups of prime degree Chandan Singh Dalawat 2011-03-19T06:40:54Z 2011-10-14T04:01:43Z <p>Happened to come across the following Satz in Huppert, <em>Endliche Gruppen I</em>, <a href="https://docs.google.com/leaf?id=0B8EHtI8F9qdIZDU0NzlkODgtZTM3MS00MGQyLWI4ZmYtMmQwMTIwYTkzNDE2&amp;hl=en_US" rel="nofollow">S. 163</a>.</p>