$p$ is the minimal prime dividing the order of $G$, and $H$ operates on $G/H$ by multiplication. $H/\ker\left(\varphi\right)$ embedded in $S_{p-1}$

Question

Let $G$ be a group, and $p$ the minimal prime the divides $\left|G\right|$. $H\leq G$ such that $\left|G:H\right|=p$. Consider the action of the group $H$ on the left cosets $G/H$ by left multiplication. It induces the homomorphism $\varphi:H\rightarrow \text{Sym}\left(G/H\right)$. Show that $H/\ker\left(\varphi\right)$ embedded in $S_{p-1}$.

I'm not sure how to approach this, I know that $\text{Sym}\left(G/H\right)\cong S_{p}$ and clearly $\left|\text{Sym}\left(G/H\right)\right|=p!$ but I'm not sure how to proceed.

Answer 1

You can prove that all such $H$ must be normal and hence the map $\phi$ is constant, so the conclusion follows trivially.

There might be something wrong in your question?

Answer 2

If a subgroup $H$ has minimal (prime) index, then it's a normal subgroup.

Moreover the action of $H$ on $G/H $ by left multiplication is given by

\begin{align} \ker(\varphi) & = \{h \in H| hgH = gH \ \text{for all}\ g \in G \} \\ & = \{h \in H| g^{-1}hgH = H \ \text{for all}\ g \in G \} \\\ \text{By normality} \ \ \ & = H \end{align}

So $\varphi$ is the trivial homomorphism.