Direct Product of Central Subgroup with Inverse Isomorphism is Central Subgroup

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Theorem

Let $G$ and $H$ be groups.

Let $\map Z G$ denote the center of $G$.

Let $Z$ and $W$ be central subgroups of $G$ and $H$ respectively.


Let:

$Z \cong W$

where $\cong$ denotes isomorphism.

Let such a group isomorphism be $\theta: Z \to W$.

Let $X$ be the set defined as:

$X = \set {\tuple {x, \map \theta x^{-1} }: x \in Z}$


Then $X$ is a central subgroup of $G \times H$.


Proof

First note that by Group Homomorphism Preserves Inverses:

$\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1} = \map \phi x^{-1}$

and so there is no amphiboly in the notation used.


It is established that $\tuple {x, \map \theta x^{-1} } \in G \times H$:

$x \in G$
$\map \theta x \in H$ and so $\map \theta x^{-1} \in H$.

Let:

$e_G$ be the identity element of $G$
$e_H$ be the identity element of $H$


We have that $e_G \in Z$, and so:

\(\displaystyle \tuple {e_G, \map \theta {e_G}^{-1} }\) \(=\) \(\displaystyle \tuple {e_G, \map \theta {e_G} }\) as $e_G^{-1} = e_G$
\(\displaystyle \) \(=\) \(\displaystyle \tuple {e_G, e_H}\) Group Homomorphism Preserves Identity

Thus $\tuple {e_G, e_H} \in X$ and so $X \ne \O$.


Then we note that:

\(\displaystyle \tuple {x, \map \theta x^{-1} } \tuple {x^{-1}, \map \theta x}\) \(=\) \(\displaystyle \tuple {x x^{-1}, \map \theta x^{-1} \map \theta x}\) Definition of Group Direct Product
\(\displaystyle \) \(=\) \(\displaystyle \tuple {e_G, e_H}\) as $\theta$ is a group isomorphism
\(\displaystyle \) \(=\) \(\displaystyle \tuple {x^{-1} x, \map \theta x \map \theta x^{-1} }\) as $\theta$ is a group isomorphism
\(\displaystyle \) \(=\) \(\displaystyle \tuple {x^{-1}, \map \theta x} \tuple {x, \map \theta x^{-1} }\) Definition of Group Direct Product

and so $\tuple {x^{-1}, \map \theta x}$ is the inverse of $\tuple {x, \map \theta x^{-1} }$ in $H$.


Let $x, y \in G$.

Then $x^{-1} \in G$ from Group Axiom $G \, 0$: Closure.


Hence if:

$\tuple {x, \map \theta x^{-1} } \in X$

it follows that:

$\tuple {x^{-1}, \map \theta x} \in X$

Let $x, y^{-1} \in G$.

Then:

$\tuple {x, \map \theta x^{-1} } \in X$

and:

$\tuple {y^{-1}, \map \theta y} \in X$


Thus:

\(\displaystyle \tuple {x, \map \theta x^{-1} } \tuple {y^{-1}, \map \theta y}\) \(=\) \(\displaystyle \tuple {x y^{-1}, \map \theta x^{-1} \map \theta y}\) Definition of Group Direct Product
\(\displaystyle \) \(=\) \(\displaystyle \tuple {x y^{-1}, \map \theta {x^{-1} y} }\) as $\theta$ is a group isomorphism
\(\displaystyle \) \(=\) \(\displaystyle \tuple {x y^{-1}, \map \theta {\paren {y^{-1} x}^{-1} } }\) Inverse of Group Product
\(\displaystyle \) \(=\) \(\displaystyle \tuple {x y^{-1}, \map \theta {\paren {x y^{-1} }^{-1} } }\) $x$ and $y$ are in the center of $G$, so commute
\(\displaystyle \) \(=\) \(\displaystyle \tuple {x y^{-1}, \map \theta {x y^{-1} }^{-1} }\)

So:

$\tuple {x, \map \theta x^{-1} } \tuple {y^{-1}, \map \theta y} \in X$

and it follows from the One-Step Subgroup Test that $X$ is a subgroup of $G \times H$.

$\Box$


We have that $Z$ is a central subgroup of $G$.

We also have that the image of $\theta$ is $W$.

From Direct Product of Central Subgroups, $Z \times W$ is a central subgroup of $G \times H$.

It follows that $X$ is a central subgroup of $G \times H$.

$\blacksquare$


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