Proving $\phi$ is a homomorphism and that the group G is simple.

Question

Let $G,H$ be groups where $H$ is a subgroup of $G$ and $[G:H] = 3$, let $g_{1},g_{2},g_{3}$ be the elements of the group $G$. Let $c_{1} = g_{1}H,c_{2} = g_{2}H,c_{3} = g_{3}H$ be three distinct left cosets of $H$ in $G$. That is if $g \in G$, we have $\phi(g) = \begin{bmatrix} c_{1} & c_{2} & c_{3} \\ gc_{1} & gc_{2} & gc_{3}\end{bmatrix}$ is a permutation. I am thinking how to show that

$\phi : G \rightarrow S_{3}$ is a homomorphism? I am thinking that $|G| = ?$ and $|S_{3}| = 6$.

Also how do we show that if $|G|$ divides 6, then $G$ is simple. I can think that the kernel is a normal subgroup whose order must divide the order of the group G.

Answer 1

Here is an answer based on my best interpretation of your question.

The group $G$ acts on $G/H$. This is given by by $$ g \cdot (g'H) = (gg')H $$

You should be able to show this explicitly. This gives a homomorphism from $G \rightarrow S_3$ because $S_3$ is the group of bijections of $G/H$ (it is a fact that an action of a group $G$ on a set $S$ gives a homomorphism $G \rightarrow \text{Aut}_\mathsf{Set}(S))$.

I don't think the claim that if $|G|$ divides $6$ then $G$ is simple is correct. For example, you can take $G$ to be $S_3$ itself.

Answer 2

Your homomorphism $\phi$ has as a kernel $ker(\phi)=\cap_{g \in G}H^g$, so the intersection of all $G$-conjugates of $H$. This is a normal subgroup of $G$. So if $G$ is simple, this kernel becomes trivial and $\phi$ injective. This implies then that $|G| \mid 6$, since by the first isomorphism theorem $G/ker(\phi) \cong \phi[G]$ is (isomorphic to) a subgroup of $S_3$.

On the other hand, if $G$ is simple and $|G:H|=3 \mid |G| \mid 6$, then $G \cong C_3$ and $H=1$.