I've come across this problem (I'm sorry if it's poorly written but it's a translation); for a $p$-Sylow I mean a $p$-subgroup that is not contained in any other $p$-subgroup:
Let $G$ be a locally finite group with a finite $p$-Sylow. Show that all the $p$-Sylow of $G$ are finite and conjugate.
Now, the problem above this one, whose demonstration nearly drove me crazy, states that if $G$, not necessarily finite, has a finite number of conjugate for a $p$-Sylow, than all the $p$-Sylow are conjugate. Thus if in the problem I proposed I manage to show that there is a $p$-Sylow with a finite orbit under conjugation, I'm done.
The problem is that I should prove it using elementary tools (the only thing I managed to find relevant is a post on MO concerning higher level results); that said, this is the only thing I could come up with:
Let $S_0$ be the set of all the "linearly independent" (in the sense that no element can be expressed only using the other ones) elements of $G$ whose order is a power of $p$. Now let $S$ be the subgroup of $G$ generated by all the $s_i$'s of $S_0$. Now if let's say $s_1$ and $s_2$ generate a $p$-Sylow, then there can't be no $s_j$ in $S_0$ so that $\langle s_1,s_2,s_i \rangle$ is a p-group as that would mean that $\langle s_1,s_2 \rangle$ is not maximal.
Another idea was to work with conjugation on either $S$ or $S_0$, but as they're infinite I don't know how to procede
The fact is that I could show that the result is true in $S$, but I'm not sure that's true (I'm quite sure it's false), or I could work in some way in $S$ and then jumping back on $G$
Best Answer
Proposition Let 𝐺 be a locally finite group with a finite Sylow $p$-subgroup. Then all the Sylow $p$-subgroups of $G$ are finite and conjugate.
Proof Let $P$ be an arbitrary finite $p$-subgroup of $G$ and $S$ the given finite Sylow $p$-subgroup. Since $G$ is locally finite, $H=\langle P,S \rangle$ is finite. Of course, $S$ is a Sylow $p$-subgroup of $H$, hence $P$ must be contained in some ($H$-)conjugate of $S$. Hence $|P| \leq |S|$ and, by definition (in the infinite case as a maximal $p$-subgroup), the order of any $p$-subgroup of $G$ cannot be larger than $|S|$. It follows that any Sylow $p$-subgroup must be finite. And the above reasoning shows that in fact any Sylow $p$-subgroup of $G$ must be conjugate to $S$.