Theorem: Preimage of Normal Subgroup under Epimorphism is Normal Subgroup

Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups.
Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a group epimorphism.
Let $H$ be a normal subgroup of $\struct {G_2, *}$.
Then:
:$\phi^{-1} \sqbrk H$ is a normal subgroup of $\struct {G_1, \circ}$
where $\phi^{-1} \sqbrk H$ denotes the preimage of $H$ under $\phi$.

Proof:
Recall that as $\phi$ is an group epimorphism, it is also {{afortiori}} a group Homomorphism.
Hence:
:$\forall a, b \in G_1: \map \phi {a \circ b} = \map \phi a * \map \phi b$
Let $H$ be a normal subgroup of $\struct {G_2, *}$.
From Preimage of Subgroup under Epimorphism is Subgroup:
:$\phi^{-1} \sqbrk H$ is a subgroup of $\struct {G_1, \circ}$.
It remains to be shown that $\phi^{-1} \sqbrk H$ is normal.
Let $K = \phi^{-1} \sqbrk H$.
That is:
:$H = \map \phi K$
To prove that $K$ is normal, we are to show that:
:$\forall g \in G: k \in K \iff g \circ k \circ g^{-1} \in K$
and:
:$\forall g \in G: k \in K \iff g^{-1} \circ k \circ g \in K$
So, let $g \in G$ and $k \in K$ be arbitrary.
By definition of preimage of $H$ under $\phi$:
:$\exists h \in H: \map \phi k = h$
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {g \circ k \circ g^{-1} }
      | r = \map \phi g * \map \phi k * \map \phi {g^{-1} }
      | c = {{Defof|Group Homomorphism}}
}}
{{eqn | r = \map \phi g * h * \map \phi {g^{-1} }
      | c = as $h = \map \phi k$
}}
{{eqn | o = \in
      | r = H
      | c = as $H$ is normal
}}
{{eqn | ll= \leadsto
      | l = g \circ k \circ g^{-1}
      | o = \in
      | r = K
      | c = {{Defof|Preimage of Subset under Mapping}}: $K = \phi^{-1} \sqbrk H$
}}
{{end-eqn}}
In the same way:
{{begin-eqn}}
{{eqn | l = \map \phi {g^{-1} \circ k \circ g}
      | r = \map \phi {g^{-1} } * \map \phi k * \map \phi g
      | c = {{Defof|Group Homomorphism}}
}}
{{eqn | r = \map \phi {g^{-1} } * h * \map \phi q
      | c = as $h = \map \phi k$
}}
{{eqn | o = \in
      | r = H
      | c = as $H$ is normal
}}
{{eqn | ll= \leadsto
      | l = g^{-1} \circ k \circ g
      | o = \in
      | r = K
      | c = {{Defof|Preimage of Subset under Mapping}}: $K = \phi^{-1} \sqbrk H$
}}
{{end-eqn}}
Thus:
:$\forall g \in G: k \in K \implies g \circ k \circ g^{-1} \in K$
and:
:$\forall g \in G: k \in K \implies g^{-1} \circ k \circ g \in K$
{{qed|lemma}}
Then:
{{begin-eqn}}
{{eqn | l = g \circ k \circ g^{-1}
      | o = \in
      | r = K
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi {g \circ k \circ g^{-1} }
      | o = \in
      | r = H
      | c = Definition of $K$
}}
{{eqn | ll= \leadsto
      | l = \map \phi g * \map \phi k * \map \phi {g^{-1} }
      | o = \in
      | r = H
      | c = {{Defof|Group Homomorphism}}
}}
{{eqn | ll= \leadsto
      | l = \map \phi k
      | o = \in
      | r = H
      | c = as $H$ is normal
}}
{{eqn | ll= \leadsto
      | l = k
      | o = \in
      | r = K
      | c = {{Defof|Preimage of Subset under Mapping}}: $K = \phi^{-1} \sqbrk H$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = g^{-1} \circ k \circ g
      | o = \in
      | r = K
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi {g^{-1} \circ k \circ g}
      | o = \in
      | r = H
      | c = Definition of $K$
}}
{{eqn | ll= \leadsto
      | l = \map \phi {g^{-1} } * \map \phi k * \map \phi g
      | o = \in
      | r = H
      | c = {{Defof|Group Homomorphism}}
}}
{{eqn | ll= \leadsto
      | l = \map \phi k
      | o = \in
      | r = H
      | c = as $H$ is normal
}}
{{eqn | ll= \leadsto
      | l = k
      | o = \in
      | r = K
      | c = {{Defof|Preimage of Subset under Mapping}}: $K = \phi^{-1} \sqbrk H$
}}
{{end-eqn}}
Thus:
:$\forall g \in G: k \in K \impliedby g \circ k \circ g^{-1} \in K$
and:
:$\forall g \in G: k \in K \impliedby g^{-1} \circ k \circ g \in K$
{{qed|lemma}}
Putting this together:
:$\forall g \in G: k \in K \iff g \circ k \circ g^{-1} \in K$
and:
:$\forall g \in G: k \in K \iff g^{-1} \circ k \circ g \in K$
and the result follows.
{{qed}}