Preimage of Normal Subgroup under Epimorphism is Normal Subgroup

Theorem
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 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:

In the same way:

Thus:


 * $\forall g \in G: k \in K \implies g \circ k \circ g^{-1} \in K$

and:
 * $\forall g \in G: k \in K \implies g^{-1} \circ k \circ g \in K$

Then:

and:

Thus:
 * $\forall g \in G: k \in K \impliedby g \circ k \circ g^{-1} \in K$

and:
 * $\forall g \in G: k \in K \impliedby g^{-1} \circ k \circ g \in K$

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.