Here is an algebra qualifying exam problem:
Let $G$ be a finite group acting transitively on a set $X$ with cardinality $p^m$ for
some prime $p$ and nonnegative integer m. Show that any $p$-Sylow subgroup of $G$
acts transitively on $X$.
There are some easy facts from the orbit-stabilizer theorem (a $p$-Sylow subgroup has order at least $p^m$, for example), but I don't see an obvious way to connect this to the $P$-orbits.
How can I solve this?
Best Answer
From the orbit-stabilizer theorem we have $|G| = p^m |S_G(x)|$, where $S_G(x)$ is the stabilizer in $G$ of any $x \in X$. Write $|S_G(x)| = |G| / p^m = p^k \cdot n$, where $n$ is coprime with $p$. Then any $p$-Sylow subgroup has order $p^{m + k}$.
Let $P$ be a $p$-Sylow subgroup of $G$. Again, by orbit-stabilizer, the $P$-orbit of an element $x \in X$ has cardinality $$\frac{|P|}{|S_P(x)|} = \frac{p^{m + k}}{|S_P(x)|}.$$ Since $S_P(x)$ is a subgroup of $S_G(x)$ and $P$, we have $|S_P(x)| \leq p^k$, thus the orbit has size at least $$\frac{p^{m + k}}{p^k} = p^m = |X|,$$ so $P$ acts transitively on $X$.
(This is basically a nice idea tucked away in an old comment by Derek Holt. I thought it deserved to be its own answer somewhere.)