Technicality of Sylow's second theorem

Question

If $P_{1}$ and $P_{2}$ are Sylow $p$-subgroups of $G$ then there exists $g\in{G}$ such that $P_{2}=gP_{1}g^{-1}$. Since $G$ could contain more elements than there are Sylow $p$-subgroups is it assumed that there could exist more than one $g\in{G}$ such that $P_{2}=gP_{1}g^{-1}$ and hence the cardinality of the orbit of $P_{1}$, where $G$ acts on the Sylow $p$-subgroups by conjugation, is equal to the number of Sylow $p$-subgroups?

Answer 1

Yes, the for the action of $G$ by conjugation on subgroups, one orbit contains exactly the Sylow $p$-subgroups, because any two Sylow $p$-subgroups of $G$ are conjugate in $G$.

This has nothing to do with the fact that there is, in general, more than one element $g$ that conjugates a given Sylow $P_1$ into $P_2$. Recall that $gKg^{-1} = hKh^{-1}$ if and only if $h^{-1}g\in N_G(K)$. And since every subgroup normalizes itself, it follows that for every $x\in P_1$ (and possibly for more, if $P_1$ is properly contained in its normalizer), you have $gxP_1(gx)^{-1}=gP_1g^{-1} = P_2$. By the Orbit-Stabilizer Theorem, we know that the number of Sylow $p$-subgroups of $G$ is equal to $[G:N_G(P)]$, where $P$ is any Sylow $p$-subgroups of $G$. In fact, that is how the clause of the Third Sylow Theorem that establishes the number divides $|G|$ is usually proven.

Answer 2

• "more than one $g \in G$ such that $P_2 = g P_1 g^{=1}$.": This is automatic. Let $g$ be such a $g$ and $p \in P_1$ (we may even assume $p$ is not the identity). Then $(gp)P_1(gp)^{-1} = g P_1 g^{-1} = P_2$. This follows from the elementary fact that $P_1$ is closed under multiplication.
• "the cardinality of the orbit of $P_1$ [under the conjugation action of $G$] is equal to the number of Sylow $p$-subgroups": This is not automatic. The orbit is automatically a subset of the Sylow $p$-subgroups. Showing that the orbit is all of the Sylow $p$-subgroups, i.e., that the conjugation action of $G$ on the Sylow $p$-subgroups is transitive, is usually a part of the third Sylow theorem.