Nucleus, Kernel in Abstract Algebra 📂Abstract Algebra

Nucleus, Kernel in Abstract Algebra

Definition

The kernel of $G, G'$ with respect to the identity element $e, e'$ and the homomorphism $\phi : G \to G'$ is the preimage $\phi^{-1} [ \left\{ e' \right\} ]$ of $\left\{ e' \right\}$ and is denoted 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.

Description

Theorem [3] is a necessary and sufficient condition, but is particularly useful in proving that a homomorphism is injective. In linear algebra, the null space is strongly identified as the solution set for a given equation.

Meanwhile, in abstract algebra, at least in group theory, $G$ is strongly characterized as ‘holding the center’, regardless of what it is, by becoming a normal subgroup. Interestingly, in theorem [1], it doesn’t even bother with how $\phi$ is actually defined or what group $G'$ is. The theorem only views $G'$ as receiving $\phi$ from $G$ , meaningless otherwise.

Proof

[3]

If $( \implies )$ $\ker \phi = \left\{ e \right\}$ , then for every $g \in G$ , $\phi ( \left\{ g \right\} )$ corresponds precisely only to $\left\{ g \right\} = g \left\{ e \right\}$ , therefore $\phi$ is injective.

Since $( \impliedby )$ $\phi$ is injective and due to $\phi (e) = e'$ , $\ker \phi = \left\{ e \right\}$ must be followed.

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