Let $G$ be a group of order $56$, $Q$ a normal Sylow $2$-group and $P$ a Sylow $7$-group. Show either $G \cong Q \times P$ or $Q \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.
The hint given is to note that $P$ acts on $Q \setminus \{e\}$ by conjugation and to show that this action is either trivial or transitive.
I understand the hint; if the action is trivial, then the elements of $Q$ and $P$ commute with each other, which means $QP$ is an internal direct sum with $56$ elements, whence $G = QP \cong Q \times P$. If the action is transitive, we know that $Q$ has some element of order $2$; therefore every element of $Q \setminus \{e\}$ has order $2$ and $Q$ is isomorphic to the given group.
The set $Q\setminus\{e\}$ has 7 elements. Let $S \leq P$ be the set of elements of $P$ that commute with some specific $q \in Q\setminus\{e\}$. If $S=\{e\}$, then by the orbit-stabilizer theorem, the orbit of $q$ under the action of $P$ has size $[P:S]=7$. In other words, $Q\setminus\{e\}$ is itself the orbit of $q$, and so $P$ acts transitively on $Q\setminus \{e\}$. This is true no matter which $q \in Q\setminus\{e\}$ is chosen, as long as $S=\{e\}$. Hence we may assume that $S \neq \{e\}$ for any $q \in Q\setminus \{e\}$ (the only remaining case). However, the only possibility for $S$ other than $\{e\}$ is $S=P$, since $|P|=7$ (Lagrange's theorem). In other words, the only remaining case is where every element of $P$ commutes with every element of $Q \setminus \{e\}$ (and obviously also with every element of $\{e\}$). This exactly means the action of $P$ on $Q \setminus\{e\}$ is trivial, and also that $G=P \times Q$.