Pullback of Quotient Group Isomorphism is Subgroup

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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:

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

if and only if:

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

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.

$\Box$


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:

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

$\blacksquare$


Sources