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:

\(\ds \map \theta N\)

\(=\)

\(\ds K\)

\(\ds \map \theta {x N}\)

\(=\)

\(\ds y K\)

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.

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $13$: Direct products: Example $13.12$