Kernel is Trivial iff Monomorphism/Group

Theorem
Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be a group homomorphism.

Let $\ker \left({\phi}\right)$ be the kernel of $\phi$.

Then $\phi$ is a group monomorphism $\ker \left({\phi}\right)$ is trivial.

Necessary Condition
Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be a group monomorphism.

By Homomorphism to Group Preserves Identity, $e_S \in \ker \left({\phi}\right)$.

If $\ker \left({\phi}\right)$ contained another element $s \ne e_S$, then $\phi \left({s}\right) = \phi \left({e_S}\right) = e_T$ and $\phi$ would not be injective, thus not be a group monomorphism.

So $\ker \left({\phi}\right)$ can contain only one element, and that must be $e_S$, which is therefore the trivial subgroup of $S$.

Sufficient Condition
Now suppose $\ker \left({\phi}\right) = \left\{{e_S}\right\}$.

Then, for $x, y \in S$:

Thus $\phi$ is injective, and therefore a group monomorphism.