A question on a proof that $A_5$ has no subgroup of order $20$ 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?

Suppose $H$ is such a subgroup. The action of $A_5$ to the left cosets $A_5/H$ induces a homomorphism $\rho: A_5 \to S_3$ with $\boxed{\ker \rho \subseteq H}$. From that we deduce that $\ker \rho = \{1\}$ (due to $A_5$ being simple) and then $5!$ would divide $6$, a contradiction.
