Group Epimorphism Preserves Normal Subgroups

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Theorem

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.

Let $\phi: G \to H$ be a group epimorphism.

Let $N \lhd G$, where $\lhd$ denotes that $N$ is a normal subgroup of $G$.


Then $\phi \sqbrk N \lhd H$.


That is, the image under $\phi$ of a normal subgroup is itself normal.


Proof

Let $N' := \phi \sqbrk N$.

From Group Homomorphism Preserves Subgroups, $N'$ is a subgroup of $H$.

It remains to show that $N'$ is normal in $H$.


Let $h \in H$ be arbitrary.

Let $n' \in N'$ be arbitrary.


Because $\phi$ is an epimorphism, it is by definition surjective.

Therefore:

$\exists n \in N: \map \phi n = n'$
$\exists g \in G: \map \phi g = h$

Then:

\(\ds g \circ n \circ g^{-1}\) \(\in\) \(\ds N\) Definition of Normal Subgroup
\(\ds \leadsto \ \ \) \(\ds \map \phi {g \circ n \circ g^{-1} }\) \(\in\) \(\ds N'\) as $N'$ is the image of $N$
\(\ds \leadsto \ \ \) \(\ds \map \phi g * \map \phi n * \map \phi {g^{-1} }\) \(\in\) \(\ds N'\) as $\phi$ has the morphism property
\(\ds \leadsto \ \ \) \(\ds \map \phi g * \map \phi n * \paren {\map \phi g}^{-1}\) \(\in\) \(\ds N'\) Group Homomorphism Preserves Inverses
\(\ds \leadsto \ \ \) \(\ds h * n' * h^{-1}\) \(\in\) \(\ds N'\) Definition of $g$ and $n$


As $h$ and $n'$ were arbitrary, it follows that:

$\forall h \in H, n' \in N': h * n' * h^{-1} \in N'$

and so $N'$ is a normal subgroup of $H$.

$\blacksquare$


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