Statement of Sylow's Fourth Theorem (single conjugacy class) I am confused by the statement of Sylow's Fourth Theorem:
Let $G$ be a finite group, $p$ a prime. The Sylow $p$-subgroups of $G$ form a single conjugacy class of subgroups.
In particular, I do not understand what it means for the subgroups to form a single conjugacy class?
Thanks!
 A: Given a group $G$, you can define an equivalence relation $\sim$ on the set of all subgroups of $G$, where if $H$ and $K$ are subgroups of $G$, 
$$H\sim K\iff \text{there exists some }g\in G\text{ such that }K=gHg^{-1}$$
In other words, $H\sim K$ if and only if $K$ is a conjugate of $H$.
Given a subgroup $H$ of $G$, the conjugacy class of $H$ is just the equivalence class of $H$, i.e.,
$$\{\text{subgroups }K\text{ of }G\mid H\sim K\}$$
So to say that the Sylow $p$-subgroups of $G$ form a single conjugacy class of subgroups of $G$ just means that there is some $H$ such that
$$\{\text{Sylow $p$-subgroups of $G$}\}=\{\text{subgroups }K\text{ of }G\mid H\sim K\}$$
It is a general fact/definition that


*

*everything equivalent to a given element of an equivalence class is again an element of the equivalence class

*every element of an equivalence class is equivalent to every other element.


Thus, every Sylow $p$-subgroup of $G$ is conjugate to every other Sylow $p$-subgroup of $G$, and moreover every conjugate of a given Sylow $p$-subgroup is another Sylow $p$-subgroup.
A: The Theorem is saying that given two Sylow $p$-subgroups $H$ and $H'$ of the group $G$, then  there exists an element $g\in G$ such that $H=gH'g^{-1}$.  Moreover, given any $g\in G$ and any Sylow $p$-subgroup $H$, then $gHg^{-1}$ is another Sylow $p$-subgroup.
That is, every pair of Sylow $p$-subgroups are conjugate subgroups, and so when you partition the collection of all subgroups into conjugacy classes, every Sylow $p$-subgroup lies in a common conjugacy class, and that conjugacy class only contains Sylow $p$-subgroups.
