Kernel is Trivial iff Monomorphism/Group

Theorem
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a group homomorphism.

Let $\map \ker \phi$ be the kernel of $\phi$.

Then $\phi$ is a group monomorphism $\map \ker \phi$ is trivial.

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

By Homomorphism to Group Preserves Identity, $e_S \in \map \ker \phi$.

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

So $\map \ker \phi$ can contain only one element, and that must be $e_S$, which is therefore the trivial subgroup of $S$.

Sufficient Condition
Now suppose $\map \ker \phi = \set {e_S}$.

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

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