(Generalization) A subgroup of smallest prime index where p dives |G|, is normal

The next theorem is a generalization of the previous theorem that a subgroup of index 2 is normal.

$\textit{Corollary 5} \quad $ (D+F p. 120): Let $H < G$. If $|G:H|=p$ where $p$ is the smallest prime divisor of $G$, then $H \lhd G$.

$\textit{Proof:}\quad$ Let ${G/H}^{\star}$ represent the set of left cosets of $H$ in $G$ (which is not necessarily a group since we have not yet established that $H$ is normal). Define a group action as a map $$\varphi: G \times {G/H}^{\star} \rightarrow {G/H}^{\star} $$ Group actions induce homomorphisms, and in this case, the induced homomorphism is $$\phi: G \rightarrow S_{|{G/H}^{\star}|} \cong S_p$$ By the first isomorphism theorem, $ker\phi \unlhd H < G$. If we can show that $ker\phi = H$, then $H \lhd G$ as desired.
Say $|H:ker\phi|=k$.
$$|G:ker\phi|=|G:H|\cdot |H:ker\phi|=pk$$
Also from the first isomorphism theorem, $G/\ker\phi \cong im\phi \lesssim S_p $. So, $$pk|p!$$ Dividing by $p$, $$k|(p-1)!$$ But, by assumption, $p$ is the smallest prime divisor, so this forces $k=1$. $\Longrightarrow ker\phi =H \therefore H \lhd G \quad \blacksquare$