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:

$\tuple {e_G, e_H} \in G \times^\theta H$
$\map \theta {e_G \circ N} = e_H * K$

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

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

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

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

$\Box$

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:

 $\displaystyle \paren {h * h'} * K$ $=$ $\displaystyle h * K * h' * K$ $\displaystyle$ $=$ $\displaystyle \map \theta {g \circ N} * \map \theta {g' \circ N}$ $\displaystyle$ $=$ $\displaystyle \map \theta {\paren {g \circ g'} \circ N}$

Thus:

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

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

$\Box$

Condition $(3)$

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

Then:

 $\displaystyle \map \theta {g \circ N}$ $=$ $\displaystyle h * K$ $\displaystyle \leadsto \ \$ $\displaystyle \map \theta {g \circ N}^{-1}$ $=$ $\displaystyle h^{-1} * K$

Then:

 $\displaystyle \map \theta {g \circ N}^{-1}$ $=$ $\displaystyle \map \theta {g^{-1} \circ N}$ Group Homomorphism Preserves Inverses $\displaystyle \leadsto \ \$ $\displaystyle \map \theta {g^{-1} \circ N}$ $=$ $\displaystyle h^{-1} * K$ $\displaystyle \leadsto \ \$ $\displaystyle \tuple {g^{-1}, h^{-1} }$ $\in$ $\displaystyle G \times^\theta H$

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

$\Box$

Therefore by the Two-Step Subgroup Test:

$G \times^\theta H \le G \times H$

$\blacksquare$