So, I'm having some problems with understanding one part of the proof of Sylow's Theorem.
Let $G$ be a group of order $n$=$p^{a}m$, where $p$ is prime and doesn't divide $m$. Then: (a) $G$ has a subgroup of order $p^{a}$....
So, here is the proof I found in Cameron's Permutation groups book:
To show (a), we let $\Omega$ be the set of all subsets of $G$ of cardinality $p^{a}$. Then $\Omega$ is a set of cardinality $\binom{n}{p^a}$. We define an action of $G$ on $\Omega$ by right multiplication: $\mu(S,g)=Sg$ for $S\in\Omega$ $g\in G$. Then $\Omega$ is the union of the $G$-orbits of this action. Note that each $G$-orbit, as a set of subsets of $G$, covers all of $G$: for if $x\in S$, then $y\in Sx^{-1}y$. So there are two types of orbit: Type 1:orbits containing exactly $n/p^{a}=m$ sets, covering $G$ without overlapping; Type 2:orbits containing more than $m$ sets. Now if $S$ is in Type 1 orbit, then its stabiliser has order $n/m=p^a$, and so is a subgroup $P$ of the required order.
The next part is a problem for me:
On the other hand, if $S$ is in a Type 2 orbit $\Delta$, then $|\Delta|>m$, so $|\Delta|$ is divisible by $p$. So, if we can show that $|\Omega|$ is not divisible by $p$, then it will follow that there must be Type 1 orbits, and (a) will be proved.
So, first, I don't get why is $|\Delta|$ divisible by $p$. Actually I don't get nothing from that problematic part so I would be thankful if someone could give me a hint or something.
Thanks!!