Pullback of Quotient Group Isomorphism/Examples/Subgroups of Index 2

Example of Pullback of Quotient Group Isomorphism
Let $G$ and $H$ be groups.

Let $N$ and $K$ be normal subgroups of $G$ and $H$ respectively such that:
 * their quotient groups $G / N$ and $H / K$ are isomorphic
 * their indices are $2$:
 * $\index G N = \index H K = 2$

Let $\theta: G / N \to H / K$ be an isomorphism.

The pullback of $G$ and $H$ by $\theta$ is a subset of $G \times H$ of the form:
 * $G \times^\theta H = \set {\tuple {g, h}: \paren {g \in N, h \in K} \text { or } \paren {g \notin N, h \notin K} }$

Proof
As $\index G N = \index H K = 2$, it follows that:
 * $\order {G / N} = \order {H / K} = 2$

and they are the cyclic group of order $2$.

Let:
 * $x \in G: x \notin N$
 * $y \in H: y \notin K$

Then:
 * $G / N = \gen {x N}$
 * $H / K = \gen {y K}$

and we have:

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

By definition:


 * $G \times^\theta H = \set {\tuple {g, h}: \map \theta {g N} = h K}$

Let $g \in N$.

Then $g N = N$ and so:
 * $\map \theta {g N} = h K = K$

So $g \in N \implies h \in K$.

Let $g \notin N$.

Then $g N = x N$ and so:
 * $\map \theta {g N} = h K = x K$

So $g \notin N \implies h \notin K$.

Hence the result.