Order of subgroup of simple group of order 168

Let $G$ be a simple group of order 168 and $H<G$. Using an action of $G$ on $G/H$, prove that $|H|\le24$.

By the given hint, I set the group action $\cdot : G \rightarrow G/H$ as $x \cdot (gH)=(xg)H$ and tried to proceed, but it doesn't work. Also, I tried to apply the equation $|X|= \sum_x |Gx|$, but I don't know how to apply properly.

As $G$ is simple, the group action $G\to Sym(G/H)$ must have trivial kernel.
Suppose $|H|>24$, then $[G:H]<7$ so $[G:H]\le 6$, but then $|\mathrm{Sym}(G/H)|=[G:H]!$ divides $6!$. As $G\to \mathrm{Sym}(G/H)$ has trivial kernel $|G|$ then divides $|\mathrm{Sym}(G/H)|$ so divides $6!$. You can check that $168$ does not divide $6!$.