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.
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 $p$.