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> [!proposition] Proposition. ([[kernel iff normal subgroup]])
> For a [[group homomorphism]] $\phi:G \to H$, the set $\text{ker }\phi$ is a [[normal subgroup]] of $G$.
> \
> The converse holds too: every [[normal subgroup]] of $G$ is the [[kernel]] of a [[group homomorphism|homomorphism]] — the **natural projection homomorphism** $\pi:G \to G / N$ given by $\pi(a)=aN$.[^1]
^0ce88b
> [!proof]- Proof. ([[kernel iff normal subgroup]])
> $\to$. Let $K$ denote the [[kernel of a group homomorphism|kernel]] of $\phi$. We will show it is [[conjugate characterization of normal subgroups|invariant under conjugation]]; let $g \in G$. We have $\phi(gkg^{-1})=\phi(g)\phi(k)\phi(g^{-1})=\phi(g)e_{H}\phi(g ^{-1})=\phi(g g^{-1})=\phi(e)=e_{H},$
> hence $gkg^{-1} \in \ker \phi$ as required.
> \
> $\leftarrow.$ Let $N \trianglelefteq G$ contain $n$, then $\pi(n)=nN=N=e_{H}$ so $n \in \ker \pi$.
[^1]: "The map which sends each element $a$ to the [[coset]] $aN$ for which it is a representative"
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#### References
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> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
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> ```
> [!frontlink]
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