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 if and only if $\map \ker \phi$ is trivial.

Proof

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 $S$.

$\Box$

Sufficient Condition

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

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

 $\ds \map \phi x$ $=$ $\ds \map \phi y$ $\ds \leadsto \ \$ $\ds \map \phi x * \paren {\map \phi y}^{-1}$ $=$ $\ds \map \phi y * \paren {\map \phi y}^{-1}$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds \map \phi {x \circ y^{-1} }$ $=$ $\ds e_T$ Definition of Morphism Property $\ds \leadsto \ \$ $\ds x \circ y^{-1}$ $\in$ $\ds \map \ker \phi$ Definition of Kernel of Group Homomorphism $\ds \leadsto \ \$ $\ds x \circ y^{-1}$ $=$ $\ds e_S$ by hypothesis $\ds \leadsto \ \$ $\ds x$ $=$ $\ds e_S \circ y$ Group Axiom $\text G 2$: Existence of Identity Element $\ds \leadsto \ \$ $\ds x$ $=$ $\ds y$ Definition of Identity Element

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

$\blacksquare$