Are Sylow p-subgroups in conjugate?

Question

Let $G$ be an infinite group and $p$ be a prime number.

Let $\mathscr{C}$ be a chain of p-subgroups of $G$ ordered by inclusion. Then, for every element of the union of $\mathscr{C}$ has an order $p^n$ hence it's a p-subgroup. Applying Zorn's lemma, $Syl_p(G)$ is nonempty.

This proof gives no information how Sylow p-subgroups are associated and I'm curious whether they are in some relation.

Let $H,K$ be Sylow p-subgroups of $G$.

Then, does there exist $g\in G$ such that $gHg^{-1}=K$? (Just like finite case?)

Answer 1

In general, the answer is no. For example, let $G$ be the direct product of countably infinitely many copies of the symmetric group $S_3$. Then $G$ is countable, but it has uncountably many Sylow $2$-subgroups, so they cannot all be conjugate.