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 $4$.
Now what do I do to solve this problem?