Action of $G$ on the left cosets of $H$ giving a non-trivial homomorphism

Question

If $H < A_5$ is a subgroup of index $3$, the action of $G$ on the left cosets of $H$ gives a non-trivial homomorphism
$$\underset{order \ 60}G \rightarrow \underset{order \ 6}{S_3}$$ which must have a kernel $$\{e\} \triangleleft_\neq N \triangleleft_\neq G$$ But G is simple so this is impossible.

I asumme that the action of $G$ is just a permutation on three elements hence $S_3$ but I cannot see how this will definitely be a non-trivial homomorphism. Nor can I see why this must have a non-trivial kernel

Does this method work in general or will there be cases where it doesnt work.

Answer 1

The general thing one can do is the following: Let $H\leq G$ and define an action of $G$ on the cosets of $H$ by $g.(kH) = (gk)H$ (so $g$ sends the coset containing $k$ to the coset containing $gk$).

It is easy to check that this is indeed an action of $G$, so it defines an homomorphism to $S_m$ where $m$ is the index of $H$ in $G$.

To see that this is a non-trivial action (when $H\neq G$), note that if $g\not\in H$ then $g.(eH) = gH \neq H$ so $g$ does not act as the identity on the coset $H$.

Answer 2

If the kernel is trivial, the map is injective, which cannot happen. If the homomorphism itself is trivial, then $H$ is normal, which also cannot happen.