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Let $G$ be a group. Part two of Sylow's theorem says that all Sylow $p$-groups (for a given $p$) are conjugate. So I assume that means given two Sylow $p$-groups $P$ and $Q$ there exists $g\in G$ such that $gPg^{-1}=Q$. Of course this is a different statement from: for all $g\in G$, $gPg^{-1}$ is a Sylow-$p$ group. Right? I ask because my lecturer keeps using the latter property, which does not seem to me to follow from the former. For example, to prove that if there is only one Sylow p-group (for a given $p$), that that subgroup must be normal, he says for all $g \in G$, $gPg^{-1}=P$, because $P$ is the only one. A bit confusing.

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  • $\begingroup$ It means the same thing $\endgroup$ – Daniel Feb 2 '18 at 21:38
  • $\begingroup$ The second Sylow's theorem asserts that $G$ operates transitively on Sylow $p$-subgroups — in other words, there's only one conjugation class. $\endgroup$ – Bernard Feb 2 '18 at 21:53
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If $P$ is a Sylow $p-$subgroup, then so is $gPg^{-1},$ since they have the same cardinality. So if the Sylow $p-$subgroup is unique, we must have $$gPg^{-1}=P$$ for all $g \in G.$ Hence, $P$ is normal.

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    $\begingroup$ oh duh. of course. $\endgroup$ – user85798 Feb 2 '18 at 22:01

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