This is a multi-step proof I'm trying to work through from my textbook. It doesn't seem to make sense, so I'm hoping someone here can point me in the right direction.
Suppose $G$ is a finite group with $p$ a prime and $p^e$ the maximal power of $p$ dividing the order of G. Let $X$ be the set of of all subsets of $G$ of order $p^e$.
The first part is to show that $|X|$ is prime to $p$. Letting $|G| = p^em$, There should be ${p^em \choose p^e}$ subsets of the desired size. I assume I just need to manipulate this expression a little to show it's relatively prime to p.
This second part is tripping me up: Let $G$ act on $X$ via the left regular action (the regular left action? Isn't that usually just the group multiplication?). Prove there is an $S \in X$ whose stabilizer P has order divisible by $p^e$.
Since the action is regular, it's transitive and semiregular, so there's only one orbit and the only thing that stabilizes $x \in X$ is the identity of $G$ for all $x$. So that means the only thing that could stabilize $S$ has to be the identity, so it's stabilizer $P$ has order one? Assuming I'm reading this right, It seems as though the statement they want me to prove can't be proven. Any insight would be appreciated.
(The last step is to conclude that $|P| = p^e$, but I haven't gotten there yet and want to try that on my own before I ask for help. I think once my confusion is cleared up I can prove this on my own.)
Once you show that $|X|$ is prime to $p$, just consider the action of $G$ on $X$. Split $X$ into $G$ orbits. You have that $|X| = |O_{s_1}| + \ldots + |O_{s_n}| = \frac{|G|}{|G_{s_1}|} + \ldots \frac{|G|}{|G_{s_n}|} = \frac{p^em}{|G_{s_1}|} + \ldots \frac{p^em}{|G_{s_n}|}$. Since $|X|$ is prime to $p$, then you must have at least one $|G_{s_i}|$ that contains $p^e$ because otherwise, you couldn't have $|X|$ to be prime to $p$. Now, WLOG $|G_{s_1}| = p^er$. This means that $G_{s_1}$ stablizes set $s_1$ of order $p^e$. When you have a group acting on a subset by left multiplication, $H$ stabilizes a set iff the set is a union of $H$-orbits. So you have that $s_i$ is a union of $G_{s_1}$ orbits, which in turn means that $p^e = |s_1| \ge |G_{s_1}| = p^er$. Thus $r = 1$ and $G_{s_1}$ is the group that you want.