Pullback of Quotient Group Isomorphism is Subgroup

Theorem
Let $\struct {G, \circ}$ be a group whose identity element is $e_G$.

Let $\struct {H, *}$ be a group whose identity element is $e_H$.

Let $N \lhd G, K \lhd H$ be normal subgroups of $G$ and $H$ respectively.

Let:
 * $G / N \cong H / K$

where:
 * $G / N$ denotes the quotient of $G$ by $N$
 * $\cong$ denotes group isomorphism.

Let $\theta: G / N \to H / K$ be such a group isomorphism.

Let $G \times^\theta H$ be the pullback of $G$ and $H$ via $\theta$.

Then $G \times^\theta H$ is a subgroup of $G \times H$.

Proof
This result is proved by an application of the Two-Step Subgroup Test:

Condition $(1)$
From the definition of pullback:
 * $\tuple {e_G, e_H} \in G \times^\theta H$


 * $\map \theta {e_G \circ N} = e_H * K$
 * $\map \theta {e_G \circ N} = e_H * K$

By Coset by Identity, $e_G \circ N, e_H * K$ are the identities of $G / N$ and $H / K$

From Group Homomorphism Preserves Identity:
 * $\map \theta {e_G \circ N} = e_H * K$

So $\tuple {e_G, e_H} \in G \times^\theta H$

Thus $G \times^\theta H$ is non-empty.

Condition $(2)$
Let $\tuple {g, h}$ and $\tuple {g', h'}$ be elements of $G \times^\theta H$.

It follows by definition of $\theta$ that:
 * $\map \theta {g \circ N} = h * K$

and:
 * $\map \theta {g' \circ N} = h' * K$

By the morphism property:
 * $\map \theta {g \circ N} * \map \theta {g' \circ N} = \map \theta {g \circ N \circ g' \circ N} = \map \theta {\paren {g \circ g'} \circ N}$

Hence:

Thus:
 * $\tuple {g \circ g', h * h'} \in G \times^\theta H$

Hence $G \times^\theta H$ is closed under the operation.

Condition $(3)$
Let $\tuple {g, h} \in G \times^\theta H$.

Then:

Then:

Thus $G \times^\theta H$ is closed under inverses.

Therefore by the Two-Step Subgroup Test:
 * $G \times^\theta H \le G \times H$