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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?

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    $\begingroup$ 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. $\endgroup$ – arctic tern Jan 3 '17 at 5:30
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    $\begingroup$ 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. $\endgroup$ – ml0105 Jan 3 '17 at 5:54
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As @arctictern put in a comment . . .

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.

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