# Central Product/Examples/D4 with Q

## Example of Central Product

Let $G$ be the dihedral group $D_4$ whose group presentation is:

- $G = \gen {a, b: a^4 = b^2 = e_G, a b = b a^{-1} }$

From Center of Dihedral Group $D_4$, the center of $G$ is:

- $\map Z G = \set {e_G, a^2}$

Let $H$ be the quaternion group $Q$ whose group presentation is:

- $Q = \gen {x, y: x^4 = e_H, y^2 = x^2, x y = y x^{-1} }$

From Center of Quaternion Group, the center of $H$ is:

- $\map Z H = \set {e_H, x^2}$

Let:

- $Z = \set {e_G, a^2}$
- $W = \set {e_H, x^2}$

Let $\theta: Z \to W$ be the mapping defined as:

- $\map \theta g = \begin{cases} e_H & : g = e_G \\ x^2 & : g = a^2 \end{cases}$

Let $X$ be the set defined as:

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

The central product of $G$ and $H$ via $\theta$ has $32$ elements.

## Proof

We have that every element $h$ of $H$ which is not in its center:

- is of order $4$
- is such that $h^2 = x^2$.

The set $X$ consists of:

- $X = \set {\tuple {e_G, e_H}, \tuple {a^2, x^2} }$

The central product of $G$ and $H$ via $\theta$ is:

- $\dfrac {G \times H} X$

Thus:

\(\ds \order {\dfrac {G \times H} X}\) | \(=\) | \(\ds \dfrac {8 \times 8} 2\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 32\) |

In the direct product $G \times H$:

- $\tuple {e_G, e_H}$ has order $1$ and is in $X$

\(\ds \tuple {a^2, e_H}^2\) | \(=\) | \(\ds \tuple {e_G, e_H} \in X\) | ||||||||||||

\(\ds \tuple {e_G, x^2}^2\) | \(=\) | \(\ds \tuple {e_G, e_H} \in X\) |

For all $h \in H$ such that $h$ is of order $4$:

\(\ds \tuple {a, h}^2\) | \(=\) | \(\ds \tuple {a^2, h^2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \tuple {a^2, x^2}\) | ||||||||||||

\(\ds \) | \(\in\) | \(\ds X\) |

For all $h \in H$ such that $h$ is of order $4$:

\(\ds \tuple {a^3, h}^2\) | \(=\) | \(\ds \tuple {a^2, h^2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \tuple {a^2, x^2}\) | ||||||||||||

\(\ds \) | \(\in\) | \(\ds X\) |

For all $g \in G$ such that $g$ is of order $2$:

\(\ds \tuple {g, e_H}^2\) | \(=\) | \(\ds \tuple {g^2, e_H}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \tuple {e_G, e_H}\) | ||||||||||||

\(\ds \) | \(\in\) | \(\ds X\) |

For all $g \in G$ such that $g$ is of order $2$:

\(\ds \tuple {g, x^2}^2\) | \(=\) | \(\ds \tuple {g^2, \paren {x^2}^2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \tuple {e_G, e_H}\) | ||||||||||||

\(\ds \) | \(\in\) | \(\ds X\) |

All remaining elements of $G \times H$ are of order $4$, and none of their squares is in $X$.

This needs considerable tedious hard slog to complete it.In particular: Not sure what direction this is going. The book is unclear.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $13$: Direct products: Example $13.10$