Let $ G $ be a finite group of order $ 2^4\times 3\times 7\times 13$. If $13 $-Sylow subgroup of $ G $ is not normal then $ G $ has 14 Sylow $13$-subgroups. Then $ G$ is $2 $-transitive on the set of Sylow subgroups. Is it possible for $ G $ or always $ G $ has a normal Sylow subgroup?

Any comments or hints are highly appreciated.


closed as off-topic by Anthony Quas, Denis Serre, Stefan Kohl, paul garrett, Nick Gill Oct 27 '14 at 16:34

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  • 1
    $\begingroup$ Note that if we factor out a central subgroup of order $6$ of ${\rm GL}(2,13)$ we are left with a group of the stated order which acts doubly transitively on its Sylow $13$-subgroups (but the central element of order $2$ is in the kernel of the permutation action). So strictly, there is a group of the order of the question which does not have a normal Sylow $13$-subgroup. $\endgroup$ – Geoff Robinson Oct 28 '14 at 22:50

Using magma, say, one can check that there is no $2$-transitive group of degree $14$ of that order. (In fact, the only $2$-transitive groups of degree 14 are $PSL(2,13)$, $PGL(2,13)$, $A_{14}$ and $S_{14}$.) (Note that $PGL(2,13)$ has half the required order.)

It's probably not too hard to prove this in an elementary way. (Such a group must be almost simple, etc...)

EDIT: See Geoff Robinson's comment below.

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    $\begingroup$ As commented above, the permutation action by conjugation on Sylow $13$-subgroups need not be faithful, and there is a group of the order stated in the question with $14$ Sylow $13$-subgroups. $\endgroup$ – Geoff Robinson Oct 29 '14 at 19:07

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