# All $3$-sylow subgroups of $S_4$

Show that all $3$-sylow subgroups in $S_4$ are conjugate. Number of 3-sylow subgroup is $(1+3k)$ s.t $8|(1+3k)$..so number of $3$-sylow subgroup is either $1$ or $4$. Since symmetric group $S_n$ ($n>3$) does not contain proper normal subgroup. So number of $3$-sylow subgroup is must be $4$..then what do I do to solve this problem?

• The number of $3$-sylow subgroups is not relevant. Since $3^2\nmid 4!$, all $3$-sylow subgroups have order $3$, hence are cyclic, and since $4<3+3$ they must be generated by $3$-cycles. All $3$-cycles are conjugate. – arctic tern Jan 3 '17 at 5:30
• Note that by the Sylow Theorems, all Sylow-$p$ subgroups (for a fixed prime $p$) are conjugate. That is, the action of the group $G$ on $\text{Syl}_{p}(G)$ by conjugation is transitive. – ml0105 Jan 3 '17 at 5:54