# $S_{4}$ has a sylow tower?

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.

• 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. – Thibaut Dumont Aug 31 '15 at 7:15
• @ThibautDumont Thanks, i edit by $H = \langle (1,2,3,4)(1,3) \rangle$ . – Soroush Aug 31 '15 at 7:18
• So it is generated by $(1,2,3,4)$ and $(1,3)$ or by the product of them which is $(1,4)(2,3)$ ? – Thibaut Dumont Aug 31 '15 at 7:19
• generated by $(1,2,3,4)$ and $(1,3)$ . – Soroush Aug 31 '15 at 7:22
• 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. – Thibaut Dumont Aug 31 '15 at 7:28

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 $p$.