# Show that all $3$-sylow subgroups of $S_4$ are conjugate.

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?

• 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

Because $$3^2\nmid 4!$$, every $$3$$-sylow subgroup has order $$3$$, so they must be cyclic. But $$4<3+3$$, so they must be generated by $$3$$-cycles. All $$3$$-cycles are conjugate.