In the begining of the Second Sylow Theorem proof my teacher did the following:
Let $G$ be finite a group, $P$ a Sylow $p$-subgroup of $G$ and $H$ any $p$-subgroup. Consider the following group action: $$H\times G/P \to G/P$$ $$(h,xP) \mapsto hxP$$ Let $\mathcal{O}_1, \mathcal{O}_2, ... \mathcal{O}_k$ be all the orbits. Because $p\nmid\ \# G/P$ and $\# G/P = \sum_{i = 1}^k |\mathcal{O}_i|$, we conclude that $\exists i \in \{1,...,k\}$ such that $p \nmid \ \# \mathcal{O}_i$. Let's fix that value for $i$.
Let $g \in G$ such that $\mathcal{O}_i = H\cdot gP$. Because of the Orbit-stabilizer theorem we have that: $$|H |= |H_{gP}| \cdot |H \cdot gP|$$ And because $H$ is a $p$-group we conclude that $|H \cdot gP|=1$.
I don't get how we can conclude that $|H \cdot gP|=1$ just by knowing that $H$ is a $p$-group and $|H |= |H_{gP}| \cdot |H \cdot gP|$. Why is this true?
Hint: your equation is of the form
(power of $p$) = (power of $p$) x (coprime to $p$)