Homomorphic Image of Quotient Group under Epimorphism

Theorem
Let $G_1$ and $G_2$ be groups whose identities are $e_{G_1}$ and $e_{G_2}$ respectively.

Let $\phi: G_1 \to G_2$ be a group epimorphism.

Let $K := \ker \left({\phi}\right)$ be the kernel of $\phi$.

Let $N$ be a normal subgroup of $G_1$ such that $K \subseteq N$.

Then:
 * $\dfrac {G_1} N \cong \dfrac {G_2}{\phi \left({N}\right)}$

where $\dfrac {G_1} N$ denotes the quotient group of $G_1$ by $N$.

Proof
From Group Epimorphism Preserves Normal Subgroups, $\phi \left({N}\right)$ is normal in $G_2$.

Let $N' := \phi \left({N}\right)$.

From Natural Epimorphism to Quotient Group, we construct the (group) epimorphism $q: G_2 \to \dfrac {G_2}{N'}$.

Now consider the composite mapping $q \circ \phi: G \to \dfrac {G_2}{N'}$ defined as:
 * $\forall q \in G: q \circ \phi \left({g}\right) = q \left({\phi \left({g}\right)}\right)$

From Composition of Homomorphisms, $q \circ \phi$ is a (group) homomorphism.

By definition of epimorphism, both $q$ and $\phi$ are surjections.

From Composite of Surjections is Surjection, $q \circ \phi$ is a surjection and therefore itself an epimorphism.

Let $g \in G$ such that:
 * $q \circ \phi \left({g}\right) = e_{G_2/N'}$

where $e_{G_2/N'}$ is the identity of $\dfrac {G_2}{N'}$.

Then $\phi \left({g}\right) \in \ker \left({q}\right)$.

But then by Group Epimorphism Induces Bijection between Subgroups it must be that $g \in N$.

Hence from the First Isomorphism Theorem, $\dfrac {G_2}{N'} \cong \dfrac {G_1} N$.