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A group having a Sylow tower is a finite group that possesses a Sylow tower: a normal series such that the successive quotient groups of the normal series all have orders that are powers of primes, and for each $ p $ dividing the order of $ G $ , there is a unique quotient that is a $ p $-subgroup and this group is isomorphic to a $ p $-Sylow subgroup of $ G $.

In other words, there exists a normal series: $ 1 = P_{0} \leq P_{1} \leq P_{2} \leq \cdots P_{r} = G $ such that for every $ p $ dividing the order of $ G $ , there exists a unique $ k $ such that $ P_{k}/P_{k-1} $is isomorphic to a $ p $-sylow subgroup of $ G $.

$ S_{4} $ have 24 elements, so a $ 2 $-sylow subgroup will have order $ 8 $ and a $ 3 $-sylow subgroup will have order $ 3 $. The subgroup $ H $ of $ S_{4} $ generated by $ (1,2,3,4) $ and $ (1,3) $ has order $ 8 $ and is thus a $ 2 $-sylow subgroup, may be isomorphic to $ D_{4} $. The subgroup $ S $ of $ S_{4} $ generated by $ (1,2,3) $ have order $ 3 $ an is thus $ 3 $-sylow subgroup of $ S_{4} $.

Now problem is that $ S_{4} $ have a sylow tower or no ? If show $ HS \unlhd G $ then $ S_{4} $ have sylow tower.

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  • $\begingroup$ Something is odd with your definition of $H$. Is it a cyclic group generated by the product $(1,2,3,4)(1,2,3)$? Or is it generated by $(1,2,3,4)$ and $(1,2,3)$ (which seems wrong since a group of order 8 can't have elements of order 3)? Please clarify. $\endgroup$ – Thibaut Dumont Aug 31 '15 at 7:15
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    $\begingroup$ @ThibautDumont Thanks, i edit by $ H = \langle (1,2,3,4)(1,3) \rangle $ . $\endgroup$ – Soroush Aug 31 '15 at 7:18
  • $\begingroup$ So it is generated by $(1,2,3,4)$ and $(1,3)$ or by the product of them which is $(1,4)(2,3)$ ? $\endgroup$ – Thibaut Dumont Aug 31 '15 at 7:19
  • $\begingroup$ generated by $ (1,2,3,4) $ and $ (1,3) $ . $\endgroup$ – Soroush Aug 31 '15 at 7:22
  • $\begingroup$ Try to prove $HS=G$. To do so conjugate (1,3) with (1,2,3) or with (1,2,3,4) and try to obtain all transpositions (1,2), (2,3), etc. $\endgroup$ – Thibaut Dumont Aug 31 '15 at 7:28
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Since neither $H$ nor $S$ is normal in $S_4$, $S_4$ does not have a Sylow tower. In fact $S_4$ is the smallest group that does not have a normal Sylow $p$-subgroup for any prime $p$.

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