Lagmental Vicfred

Kernels Are Normal Because Conjugation Cannot Leave Zero by Vicfred

The kernel of a homomorphism is automatically normal, which explains why kernels are exactly the subgroups one may quotient by. This is a compact note, but the quantifiers and hypotheses stay on the page.

The mathematical object

A homomorphism \(\varphi:G\to H\) packages a comparison of operations. Its kernel \(\ker\varphi\) measures collapse, while its image \(\operatorname{im}\varphi\) records the part of \(H\) actually reached.

$$ K=\{g\in G:\varphi(g)=e_H\} $$

I read the first line as input and the second as output. The symbols \(\forall\) and \(\exists\) are not interchangeable, and neither may be upgraded silently to \(\Longleftrightarrow\).

$$ gKg^{-1}=K\qquad(g\in G) $$

One explicit computation

A worked instance is useful here because it exposes every index that the compressed statement hides.

$$ \left.\begin{aligned}k\in K&\Longrightarrow\varphi(gkg^{-1})=\varphi(g)e_H\varphi(g)^{-1}\\&=e_H\end{aligned}\right\}\Longrightarrow gkg^{-1}\in K $$

Why the identity matters

The invariant statement is the one that does not depend on a convenient choice of coordinates, representatives, basis, or enumeration.

$$ \begin{aligned} \mathsf{D}\;&:\quad K=\{g\in G:\varphi(g)=e_H\},\\[5pt] \mathsf{C}\;&:\quad gKg^{-1}=K\qquad(g\in G). \end{aligned} $$

Where it can fail

The quotient notation \(G/N\) is legal only for \(N\trianglelefteq G\). A set of cosets may exist without inheriting a well-defined group multiplication.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] gKg^{-1}=K\qquad(g\in G) \end{gathered}} $$

I would use the boxed line as a reference later, while returning to the full display whenever a hypothesis becomes uncertain. That division keeps compression from becoming ambiguity.

This article was posted on Sat 14 December 2019. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.