Let G be a finite group and let p be the smallest prime divisor of $| G|.$ Then every subgroup H of index p in G is normal in G .
(In https://people.math.osu.edu/all.3/algebra/group%20theory/group1.pdf Autumn 2004, problem 4)
They say that if the kernel of $\varphi$ is trivial then $H$ is normal in $G$ but i can't figure out why that is true
Note that $\ker \phi \not = G$, as otherwise we know that $G = \ker \phi \le H$, which means that $G = H$.
I guess that's what the author thinks when he says that the kernel is non-trvial, although to be fair in my opinion when somebody say non-trivial I understand that $\ker \phi \not = \{e\}$. If you want to say that $\ker \phi \not = G$, you usually say that the kernel is proper (a proper subgroup of $G$). My best guess is that this is the author's thought and he just used a "different" language.
The fact $\ker \phi \not = G$ is important because it means that $[G:\ker \phi] \not = 1$. This is used to conlcude that $[G:\ker \phi] = p$, as otherwise note that $1$ is a number that divides the order of both $G$ and $S_p$.