My question regards the reasoning that was given in the answer to the question "Why does $A_5$ not have any subgroup of order $20$". The particular question can be found here:
Why $A_{5}$ has no subgroup of order 20?
It seems that the answer can be found by assuming that $H$ is such a subgroup, and realizing that $A_5$ acts on the set of left cosets of $H$, therefore implying that $A_5$ is isomorphic to a subgroup of $S_3$. An answer to this question states that the above is true because $A_5$ is a simple group.
My question is what does the group $A_5$ being a simple group have to do with $A_5$ being isomorphic to a subgroup of $S_3$? How does $A_5$ acting on the set of left cosets of $H$ imply this isomorphism?
Also, an a somewhat unrelated note we don't assume that the subgroup of order 20 is a normal subgroup, so we can't construct the quotient group $A_5/H$ which would have 3 elements, but how do we know then that the set of left cosets of $H$ will have precisely 3 elements?
Best Answer
More generally, let $G$ be a nontrivial finite group and $X_G$ the set of all the proper subgroups of $G$: $$X_G:=\{H\subseteq G\mid H\le G \wedge H\ne G\}.$$ If $G$ is simple, then:
$$[G:H]!\ge|G|, \space\forall H\in X_G \tag 1$$
In fact, for $H\in X_G$, the group $G$ acts by left multiplication on the left quotient set $G/H$, and this action has trivial kernel$^\dagger$. So, $G$ embeds into $S_{[G:H]}$, whence $(1)$.
The constraint $(1)$ means that finite simple groups can't have "relatively big" subgroups. As for your case, from $(1)$ follows that $A_5$ can't have subgroups of order $15$, $20$ and $30$ (this latter is ruled out by the very simplicity of $A_5$, though), because $[60:k]!<60$ for $k=15, 20, 30$.
$^\dagger$For $H\in X_G$, we get: $K:=\bigcap_{g\in G}gHg^{-1}\lneq G$, and thence $K=\{1\}$ for the simplicity of $G$.