Central Product/Examples/D4 with Q

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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:

\(\displaystyle \order {\dfrac {G \times H} X}\) \(=\) \(\displaystyle \dfrac {8 \times 8} 2\)
\(\displaystyle \) \(=\) \(\displaystyle 32\)


In the direct product $G \times H$:

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


\(\displaystyle \tuple {a^2, e_H}^2\) \(=\) \(\displaystyle \tuple {e_G, e_H} \in X\)
\(\displaystyle \tuple {e_G, x^2}^2\) \(=\) \(\displaystyle \tuple {e_G, e_H} \in X\)


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

\(\displaystyle \tuple {a, h}^2\) \(=\) \(\displaystyle \tuple {a^2, h^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {a^2, x^2}\)
\(\displaystyle \) \(\in\) \(\displaystyle X\)


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

\(\displaystyle \tuple {a^3, h}^2\) \(=\) \(\displaystyle \tuple {a^2, h^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {a^2, x^2}\)
\(\displaystyle \) \(\in\) \(\displaystyle X\)


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

\(\displaystyle \tuple {g, e_H}^2\) \(=\) \(\displaystyle \tuple {g^2, e_H}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {e_G, e_H}\)
\(\displaystyle \) \(\in\) \(\displaystyle X\)


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

\(\displaystyle \tuple {g, x^2}^2\) \(=\) \(\displaystyle \tuple {g^2, \paren {x^2}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {e_G, e_H}\)
\(\displaystyle \) \(\in\) \(\displaystyle X\)


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



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