Direct Product of Central Subgroup with Inverse Isomorphism is Central Subgroup

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:

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

Then we note that:

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.

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:

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 $G \times H$.

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 $G \times H$.