Suppose that G is a group with $|G| = mp$ where $p$ is a prime and $1 < m < p$. Prove that $G$ is not simple. The question I am trying to solve is:
Suppose that G is a group with  $|G| = mp$ where  $p$ is a prime and  $1 < m < p$. Prove that  $G$ is not simple.
We were given the hint to construct a homomorphism $G \to S_m$. 
I came across this proof, but I don't understand all of the parts of the proof. 
Proof:
Let  $H\leqslant$ G be such that  $\lvert H \rvert=p$. It is trivial that there is a homomorphism  $\phi:G\to \text{Sym}\left(G/H\right)$ by having  $\phi_g(aH)=gaH$. Moreover, one can prove that $\ker\phi\subseteq H$. Now, since  p is prime and  $\ker\phi\leqslant H$ we must have that  $\ker\phi=\{e\}$ or  $\ker\phi=H$. Suppose that  $\ker\phi=\{e\}$ then  $\text{im}(\phi)$ is a subgroup of $\text{Sym}\left(G/H\right)$ of order  $mp$ and so $ mp\mid m!$ but since  $p$ is prime and $ m<p$ this is impossible. Thus,  $H=\ker\phi$ and so  $\{e\}\triangleleft H\triangleleft G $ so that $G$ isn't simple.
My questions are why is it trivial that such a homomorphism exists, how do we show that the kernel is in $H$, and why is the kernel equal to either $\{e\}$ or $H$?
Thank you for any help.
 A: The group $G$ acts on the set of cosets of $H$ by left multiplication. This action is just permuting the cosets in some way. There are $m$ such cosets, so the action of $G$ on $G/H$ 'translates' to a homomorphism from $G$ to $S_m\cong \text{Sym}(G/H)$.
The details of why $\phi$ is actually a homomorphism. $\phi:G\to\text{Sym}(G/H)$ is given by $g\mapsto\phi_g$, where $\phi_g(aH)=gaH$. It's not too difficult to show that $\phi$ is a homomorphism: $\phi(g_1g_2)=\phi_{g_1g_2}$ and $$\phi_{g_1g_2}(aH)=g_1g_2aH=g_1(g_2aH)=g_1\phi_{g_2}(aH)=\phi_{g_1}\circ\phi_{g_2}(aH)=\phi(g_1)\phi(g_2)(aH).$$
So $\phi(g_1g_2)=\phi(g_1)\phi(g_2)$ and we do indeed have a homomorphism.
For the kernel: The kernel of $\phi$ is the set of elements of $G$ which are sent to permutations which fix all cosets of $H$ in $G$. Consider the permutations which fix $H$, i.e. $\phi_g(H)=H$, clearly this is true only for $g\in H$, so the kernel has to be a subgroup of $H$.
Remember that the order of $H$ is prime, and Lagrange's theorem states that the order of any subgroup must divide the order of that group, so since ker$(\phi)\leq H$, its order can only be 1 or $p$, so either $\{e\}$ or $H$ itself.
Hope this helps.
