Group Epimorphism Preserves Normal Subgroups

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 any arbitrary element of $H$.

Let $n' \in N'$ be any arbitrary element of $N'$.

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

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

Then:

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 $H$.