Theorem: Direct Product of Central Subgroup with Inverse Isomorphism is Central Subgroup

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:
{{begin-eqn}}
{{eqn | l = \tuple {e_G, \map \theta {e_G}^{-1} }
      | r = \tuple {e_G, \map \theta {e_G} }
      | c = as $e_G^{-1} = e_G$
}}
{{eqn | r = \tuple {e_G, e_H}
      | c = Group Homomorphism Preserves Identity
}}
{{end-eqn}}
Thus $\tuple {e_G, e_H} \in X$ and so $X \ne \O$.
Then we note that:
{{begin-eqn}}
{{eqn | l = \tuple {x, \map \theta x^{-1} } \tuple {x^{-1}, \map \theta x}
      | r = \tuple {x x^{-1}, \map \theta x^{-1} \map \theta x}
      | c = {{Defof|Group Direct Product}}
}}
{{eqn | r = \tuple {e_G, e_H}
      | c = as $\theta$ is a group isomorphism
}}
{{eqn | r = \tuple {x^{-1} x, \map \theta x \map \theta x^{-1} }
      | c = as $\theta$ is a group isomorphism
}}
{{eqn | r = \tuple {x^{-1}, \map \theta x} \tuple {x, \map \theta x^{-1} }
      | c = {{Defof|Group Direct Product}}
}}
{{end-eqn}}
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|0}}.
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:
{{begin-eqn}}
{{eqn | l = \tuple {x, \map \theta x^{-1} } \tuple {y^{-1}, \map \theta y}
      | r = \tuple {x y^{-1}, \map \theta x^{-1} \map \theta y}
      | c = {{Defof|Group Direct Product}}
}}
{{eqn | r = \tuple {x y^{-1}, \map \theta {x^{-1} y} }
      | c = as $\theta$ is a group isomorphism
}}
{{eqn | r = \tuple {x y^{-1}, \map \theta {\paren {y^{-1} x}^{-1} } }
      | c = Inverse of Group Product
}}
{{eqn | r = \tuple {x y^{-1}, \map \theta {\paren {x y^{-1} }^{-1} } }
      | c = $x$ and $y$ are in the center of $G$, so commute
}}
{{eqn | r = \tuple {x y^{-1}, \map \theta {x y^{-1} }^{-1} }
      | c = 
}}
{{end-eqn}}
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 $G \times H$.
{{qed|lemma}}
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 $G \times H$.
{{qed}}