If $H\leq G$ and $[G: H]=k$ then there exists a homomorphism $\varphi:G\longrightarrow S_k$? Suppose $H\leq G$ and $[G: H]=k$ for some positive integer $k$. How can I show there exists a homomorphism $\varphi:G\longrightarrow S_k$ such that $\textrm{ker}(\varphi)\leq H$?
Notation: Here $S_k$ is the group of permutations of the first $k$ positive integers and $[G: H]$ is the index of $H$ in $G$.
Thanks
 A: Hint: Consider the homomorphism which you obtain from the group action $\sigma(x,gH)=xgH$
What is the kernel of this action?
A: 1.  Enumerate the cosets:
$$
G/H = \{g_1 H, \ldots, g_k H\}.
$$
2.
Define the homomorphism $\varphi: G \to S_k$ by the action of $G$ on $G/H$.  In more detail, for any $g \in G$ and any $i~(1 \le i \le k)$,
$$
g \cdot g_i H = g_j H \quad \text{for some } j~(1\le j \le k).
$$
This implicitly defines a permutation $\sigma \in S_k$; i.e. $\sigma(i) = j$.  Set
$$
\varphi(g) = \sigma.
$$
3.  Check that this map is a homomorphism.  This is one of the axioms of a group action:
$$
(g' g) \cdot g_i H = g' \cdot ( g \cdot g_1 H )
$$
In our situation, if $\varphi(g) = \sigma$ and $\varphi(g') = \sigma'$, then $\varphi(g' g) = \sigma' \sigma$.
4.  Analyze the kernel.  Suppose that $g \in \ker \varphi$.  Then, $\varphi(g)$ is the identity permutation in $S_k$, or
$$
g \cdot g_i H = g_i H \quad \text{for all } i.
$$
This is equivalent to $g \in H$ (consider the identity coset $H$), so
$$
\ker \varphi \le H.
$$
A: Hint: $$\varphi:G\longrightarrow S_k= [f:(xH|x\in G)\rightarrow(xH|x\in G),f\ bijective]$$ $\varphi(g)=f:xH\rightarrow gxH$
