Second Sylow theorem's proof In the proof of the 2nd Sylow theorem, I don't understand why the fact tha $O(S)| m$ implies that there exists a suborbit with one element. This is part of the proof

Second Sylow Theorem. Let $G$ be a finite group with order
$p^a \cdot m$ and $p$ a prime such that $m$ is not divisible by $p$.
Then any $p$-subgroup $H$ of $G$ is contained in a $p$-Sylow subgroup.

Proof. Let $P$ be the set of $p$-Sylow subgroups of $G$, and
let $H$ be a $p$-subgroup of $G$, and let $|H| = p^r$, $r \le; a$. Let
$S$ a fixed $p$-Sylow subgroup and $O(S)$ its orbit under the
conjugation action. Let $H$ act on $O(S)$: $hT = hTh^{-1} \, \forall T
> \in O(S)$ : it's an action on on $O(S)$ because the conjugate of an
element is still in $O(S)$. $O(S)$ is thus partitioned into suborbits,
each one having a cardinality which divides $p^r$. The cardinality
$O(S)$ is the sum of the cardinalities of all these suborbits.
Since the cardinality $O(S)$ divides $m$ and is relatevely
prime to $p$, there exists at least one suborbit having just one
element; that is $\exists T \in O(S)$ such that $hTh^{-1} = T \quad
> \forall; h \in H$.

 A: Remember that the number of elements in each (sub)orbit must be a factor of $|H|$, i.e., a power of $p$. Can they all be of the form $p^\alpha$ with $\alpha\ge 1$?
A: (i) $|O(S)|$ is the number of distinct conjugates of the Sylow-$p$ subgroup $S$.
(ii) We ask: when two memebers of $G$ give same conjugate of $S$?
$$
\begin{align*}
\mbox{ $g,h\in G$ give same conjugate of $S$} & \Longleftrightarrow  gSg^{-1}=hSh^{-1}\\
& \Longleftrightarrow h^{-1}gS(h^{-1}g)^{-1}=S\\
& \Longleftrightarrow h^{-1}g \mbox{ normalizes } S\end{align*}
$$
The normalizer $N(S)$ of $S$ in $G$ is a subgroup of $G$; it contains $S$. Thus,
$$
\mbox{$g,h$ give same conjugate of $S$ } \Longleftrightarrow h^{-1}g\in N(S) \Longleftrightarrow gN(S)=hN(S).
$$
In other words, the number of distinct conjugates of $S$ is equal to index of normalizer of $S$ in $G$. Now,
$$S\subseteq N(S)\subseteq G$$
and index of $S$ in $G$ is $m$, so index of $N(S)$ in $G$ divides $m$; i.e. $|O(S)|$ divides $m$.

The bold-faced conclusion is true even if $S$ is not Sylow-$p$ subgroup; the arguments show that we not used any special thing about $S$.
