Preimage of Subgroup under Epimorphism is 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 subgroup of $\struct {G_2, *}$.

Then:
 * $\phi^{-1} \sqbrk H$ is a subgroup of $\struct {G_1, \circ}$

where $\phi^{-1} \sqbrk H$ denotes the preimage of $H$ under $\phi$.

Proof
Let $H$ be a subgroup of $\struct {G_2, *}$.

First note that from Null Relation is Mapping iff Domain is Empty Set:
 * $\phi^{-1} \sqbrk H = \O \implies H = \O$

But $H \ne \O$.

Hence $\phi^{-1} \sqbrk H$ is not empty.

Next, let $x, y \in \phi^{-1} \sqbrk H$.

Then:
 * $\exists h_1, h_2 \in H: h_1 = \map \phi x, h_2 = \map \phi y$

From the definition of Group Homomorphism, we have:
 * $\map \phi {x^{-1} \circ y} = h_1^{-1} * h_2$

Since $H$ is a subgroup:
 * $h_1^{-1} * h_2 \in H$

Hence:
 * $x^{-1} \circ y \in \phi^{-1} \sqbrk H$

Thus from the One-Step Subgroup Test:
 * $\phi^{-1} \sqbrk H$ is a subgroup of $G_1$.