Kernel in Abstract Algebra
Definition
For the identities $e, e'$ of $G, G'$ and a homomorphism $\phi : G \to G'$, the preimage $\phi^{-1} [ \left\{ e' \right\} ]$ of $\left\{ e' \right\}$ is called the kernel of $\phi$ and is written as $\ker \phi$.
Theorem
- [1]: For $g \in G$, $g ( \ker \phi ) = ( \ker \phi ) g$
- [2]: $\ker \phi \triangleleft G$
- [3]: $\ker \phi = \left\{ e \right\}$ $\iff$ $\phi$ is injective.
- [4]: If $\phi$ is surjective and $\ker \phi = \left\{ e \right\}$, then $\phi$ is an isomorphism.
Explanation
Theorem [3] is a necessary and sufficient condition, but it is especially useful for showing that a homomorphism is injective. In linear algebra, the null space had a strong identity as the solution set of a given equation.
In contrast, in abstract algebra, at least in group theory, its character is rather that of something that 'anchors the center' by being a normal subgroup no matter what $G$ is. What is curious is that theorem [1] does not even care how $\phi$ is actually defined or what kind of group $G'$ is. Looking at the theorem alone, $G'$ is nothing but a target that receives the $\phi$ fired from $G$, carrying no meaning of its own.
Proof
[3]
$( \implies )$ If $\ker \phi = \left\{ e \right\}$, then for every $g \in G$, $\phi ( \left\{ g \right\} )$ corresponds exactly only to $\left\{ g \right\} = g \left\{ e \right\}$, so $\phi$ is injective.
$( \impliedby )$ Since $\phi$ is injective and $\phi (e) = e'$, it must be that $\ker \phi = \left\{ e \right\}$.
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