Pullback of Quotient Group Isomorphism is Subgroup

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity element is $e_G$.

Let $\left({H, *}\right)$ 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:
 * $\left({e_G, e_H}\right) \in G \times^\theta H$


 * $\theta \left({e_G \circ N}\right) = e_H * K$
 * $\theta \left({e_G \circ N}\right) = e_H * K$

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:
 * $\theta \left({e_G \circ N}\right) = e_H * K$

So $\left({e_G, e_H}\right) \in G \times^\theta H$

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

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

It follows by definition of $\theta$ that $\theta \left({g \circ N}\right) = h * K$ and $\theta \left({g' \circ N}\right) = h' * K$.

By the morphism property:
 * $\theta \left({g \circ N}\right) * \left({g' \circ N}\right) = \theta\left({g \circ N \circ g' \circ N}\right) = \theta \left({\left({g \circ g'}\right) \circ N}\right)$

Thus, $\left({g \circ g', h * h'}\right) \in G \times^\theta H$.

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

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

Then:
 * $\theta \left({g \circ N}\right) = h * K$

so:
 * $\theta \left({g \circ N}\right)^{-1} = h^{-1} * K$

By Group Homomorphism Preserves Inverses:
 * $\theta \left({g \circ N}\right)^{-1} = \theta \left({g^{-1} \circ N}\right)$

so:
 * $\theta \left({g^{-1} \circ N}\right) = h^{-1} * K$

So:
 * $\left({g^{-1}, h^{-1} }\right) \in G \times^\theta H$.

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

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