Problem
Write out the group tables for \(S_3, D_8\), and \(Q_8\).
Solution
\(S_3\)
By definition, \(S_3\) is a group of permutations of \(\{1, 2, 3\}\). Its elements are the following.
- \(x_1 = 1\)
- \(x_2 = (1\ 2)\)
- \(x_3 = (1\ 3)\)
- \(x_4 = (2\ 3)\)
- \(x_5 = (1\ 2\ 3)\)
- \(x_6 = (1\ 3\ 2)\)
Given that \(x_1\) is the identity, the first column and the first row are trivial.
For the second row, $$ x_2^2 = x_1,\ x_2x_3 = (1\ 3\ 2) = x_6,\ x_2x_4 = (1\ 2\ 3) = x_5,\ x_2x_5 = (2\ 3) = x_4, x_2x_6 = (1\ 3) = x_3 $$
For the third row, $$ x_3x_2 = (1\ 2\ 3) = x_5,\ x_3^2 = x_1,\ x_3x_4 = (1\ 3\ 2) = x_6,\ x_3x_5 = (1\ 2) = x_2,\ x_3x_6 = (2\ 3) = x_4 $$
For the fourth row, $$ x_4x_2 = (1\ 3\ 2) = x_6,\ x_4x_3 = (1\ 2\ 3) = x_5,\ x_4^2 = x_1,\ x_4x_5 = (1\ 3) = x_3,\ x_4x_6 = (1\ 2) = x_2 $$
For the fifth row, $$ x_5x_2 = (1\ 3) = x_3,\ x_5x_3 = (2\ 3) = x_4,\ x_5x_4 = (1\ 2) = x_2,\ x_5^2 = (1\ 3\ 2) = x_6, x_5x_6 = x_1 $$
For the sixth row, $$ x_6x_2 = (2\ 3) = x_4,\ x_6x_3 = (1\ 2) = x_2,\ x_6x_4 = (1\ 3) = x_3,\ x_6x_5 = x_1,\ x_6^2 = (1\ 2\ 3) = x_5 $$
The resulting group table is as follows.
$$ \begin{pmatrix} x_1 & x_2 & x_3 & x_4 & x_5 & x_6\\x_2 & x_1 & x_6 & x_5 & x_4 & x_3\\x_3 & x_5 & x_1 & x_6 & x_2 & x_4\\x_4 & x_6 & x_5 & x_1 & x_3 & x_2\\x_5 & x_3 & x_4 & x_2 & x_6 & x_1\\x_6 & x_4 & x_2 & x_3 & x_1 & x_5 \end{pmatrix} $$
$$\tag*{$\blacksquare$}$$
\(D_8\)
By definition, \(D_{2n}=\langle r,s | r^n = s^2 = 1, rs = sr^{-1}\rangle\) and, in particular,
$$ D_8 = \{1, r, r^2, r^3, s, sr, sr^2, sr^3\} $$
The group table is
$$ \begin{pmatrix} 1 & r & r^2 & r^3 & s & sr & sr^2 & sr^3\\r & r^2 & r^3 & 1 & sr^3 & s & sr & sr^2\\r^2 & r^3 & 1 & r & sr^2 & sr^3 & s & sr\\r^3 & 1 & r & r^2 & sr & sr^2 & sr^3 & s\\s & sr & sr^2 & sr^3 & 1 & r & r^2 & r^3\\sr & sr^2 & sr^3 & s & r^3 & 1 & r & r^2\\sr^2 & sr^3 & s & sr & r^2 & r^3 & 1 & r\\sr^3 & s & sr & sr^2 & r & r^2 & r^3 & 1 \end{pmatrix} $$
$$\tag*{$\blacksquare$}$$
\(Q_8\)
Straight from the definition of quaternion group,
$$ \begin{pmatrix} 1 & -1 & i & -i & j & -j & k & -k\\ -1 & 1 & -i & i & -j & j & -k & k\\i & -i & -1 & 1 & k & -k & -j & j\\ -i & i & 1 & -1 & -k & k & j & -j\\j & -j & -k & k & -1 & 1 & i & -i\\ -j & j & k & -k & 1 & -1 & -i & i\\k & -k & j & -j & -i & i & -1 & 1\\ -k & k & -j & j & i & -i & 1 & -1 \end{pmatrix} $$
$$\tag*{$\blacksquare$}$$