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 $hTh^{-1} = T \quad
> \forall; h \in H$.


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 $\alpha\ge 1$?
