$G$ is a group of order 168 , $H$ <$G$ Proof that $[G:H]=m\geq 7$ using a morphism

Question

let $G$ be a simple group of order $168$. Let $H$ be a subgroup such that $[G:H]=m, m>1$ Considering the action of $G$ on $G/H$ by $g.xH= (gx)H$

This operation is transitive so this implies a morphism between $G$ and Sm, the symmetric group of degree $m$.

Since $G$ is simple, the morphism is injective.

I don't understand the following sentence : Hence the order of $G$ divides $m$. That is possible if and only if $m \geq 7$

If someone can help me to understand why the order of $G$ divides $m$ ? ( i agree that since the morphism is injective the order of g has to be superior to the order of Sm= m! but i don't see why it has to divides.

Answer 1

Consider the morphism $$\phi : G \to S_m$$

Since this is injective $G$ is isomorphic to $\phi(G)$ and hence, they have the same cardinality.

Since $\phi(G)$ is a subgroup of $S_m$, by Lagrange Theorem, the order of $\phi(G)$ divides the order of $S_m$, which is $m!$.