Here is the second paragraph of the proof:
"First observe that $G$ has no proper subgroup $H$ of index less than $5$, as follows: if $H$ were a subgroup of index $4$, $3$ or $2$, then, by Theorem 3, $\mathbf{G}$ would have a normal subgroup $\mathbf{K}$ contained in $\mathbf{H}$ with $\mathbf{H/K}$ isomorphic to a subgroup of $\mathbf{S_4}$, $\mathbf{S_3}$ or $\mathbf{S_2}$".
And here is Theorem 3:
"Let $G$ be a group, let $H$ be a subgroup of $G$ and let $G$ act by left multiplication on the set $A$ of left cosets of $H$ in $G$. Let $\pi_H$ be the associated permutation representation afforded by this action. Then (1) $G$ acts transitively on $A$ (2) the stabilizer in $G$ of the point $1H\in A$ is the subgroup of $H$ (3) the kernel of the action (i.e., the kernel of $\pi_H$) is $\bigcap_{x\in G}xHx^{-1}$, and $\ker\pi_H$ is the largest normal subgroup of $G$ contained in $H$."
My Question: How exactly do the boldface texts follow from Theorem 3? Part (3) of the theorem merely says that a normal group contained in $H$ exists, but does not say anything about its order. If, for example, $|H|=20$, then $|K|$ could well be $2$, and then $|G/K|=30$, thus the best can be said (with the help of Cayley's Theorem) is that $\mathbf{G/K}$ is isomorphic to a subgroup of $\mathbf{S_{30}}$. What am I missing here?
Any help would be greatly appreciated.
In Theorem 3, the action is on the set of left cosets of $H$. So the map $\pi_H$ is a homomorphism $\pi_H: G \rightarrow S_k$, where $k = [G:H]$ and $\operatorname{Ker} \pi_H \leq H$.
Then by the first isomorphism theorem $G/\operatorname{Ker} \pi_H$ is isomorphic to $\pi_H(G) \leq S_k$.