Kernel is Trivial iff Monomorphism/Group

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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 $x, y \in S$:

\(\displaystyle \map \phi x\) \(=\) \(\displaystyle \map \phi y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \phi x * \paren {\map \phi y}^{-1}\) \(=\) \(\displaystyle \map \phi y * \paren {\map \phi y}^{-1}\) Group Axioms
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \phi {x \circ y^{-1} }\) \(=\) \(\displaystyle e_T\) Definition of Morphism Property
\(\displaystyle \leadsto \ \ \) \(\displaystyle x \circ y^{-1}\) \(\in\) \(\displaystyle \map \ker \phi\) Definition of Kernel of Group Homomorphism
\(\displaystyle \leadsto \ \ \) \(\displaystyle x \circ y^{-1}\) \(=\) \(\displaystyle e_S\) by hypothesis
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle e_S \circ y = y\) Group Axioms


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

$\blacksquare$


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