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### There is at most one prime $p$ such that if $P\in \operatorname{Syl}_p(G)$, then $N_G\left ( P \right )=P.$

As a hint, let $H$ be a maximal (proper) normal subgroup of $G$, so $|G:H|=p$ is prime. Then by the lemma that you stated, $p$ is the only prime for which a Sylow $p$-subgroup of $G$ can be self-...
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### Are Sylow $p$-subgroups in infinite groups conjugate?

In addition to what has been answered: if $P$ and $Q$ are arbitrary $p$-groups, then $P$ and $Q$ are Sylow $p$-subgroups of their free product $G=P*Q$. Hence, in an infinite group Sylow $p$ subgroups ...
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