symmetric group on three letters

This example is of the symmetric group on $3$ letters, usually denoted by $S_{3}$ . Here, we are considering the set of bijective functions on the set $A=\\{1,2,3\\}$ which naturally arise as the set of permutations on $A$ . Our binary operation is function composition which results in a new bijective function. This example develops the table for $S_{3}$ . We start by listing the elements of our group. These elements are listed according to the second method as described in the entry on permutation notation.

e={1\\ 2\\ 3\\choose 1\\ 2\\ 3}\\hskip 56.905512ptr={1\\ 2\\ 3\\choose 2\\ 1\\ 3}

a={1\\ 2\\ 3\\choose 2\\ 3\\ 1}\\hskip 56.905512pts={1\\ 2\\ 3\\choose 3\\ 2\\ 1}

b={1\\ 2\\ 3\\choose 3\\ 1\\ 2}\\hskip 56.905512ptt={1\\ 2\\ 3\\choose 1\\ 3\\ 2}

Here, our group is just $S_{3}=\\{e,a,b,r,s,t\\}$ . Now we can start to multiply and then fill in the table.First, we calculate the square of each element.

a^{2}={1\\ 2\\ 3\\choose 2\\ 3\\ 1}{1\\ 2\\ 3\\choose 2\\ 3\\ 1}={1\\ 2\\ 3\\choose 3\\ 1\\ 2%}=b

b^{2}={1\\ 2\\ 3\\choose 3\\ 1\\ 2}{1\\ 2\\ 3\\choose 3\\ 1\\ 2}={1\\ 2\\ 3\\choose 2\\ 3\\ 1%}=a

r^{2}={1\\ 2\\ 3\\choose 2\\ 1\\ 3}{1\\ 2\\ 3\\choose 2\\ 1\\ 3}={1\\ 2\\ 3\\choose 1\\ 2\\ 3%}=e

s^{2}={1\\ 2\\ 3\\choose 3\\ 2\\ 1}{1\\ 2\\ 3\\choose 3\\ 2\\ 1}={1\\ 2\\ 3\\choose 1\\ 2\\ 3%}=e

t^{2}={1\\ 2\\ 3\\choose 1\\ 3\\ 2}{1\\ 2\\ 3\\choose 1\\ 3\\ 2}={1\\ 2\\ 3\\choose 1\\ 2\\ 3%}=e

Next, we will fill in the upper right $3\\operatorname{x}3$ block, we only need $a b$ and $b a$ since we can use the fact that there can be no repetition in any row or column.

ab={1\\ 2\\ 3\\choose 2\\ 3\\ 1}{1\\ 2\\ 3\\choose 3\\ 1\\ 2}={1\\ 2\\ 3\\choose 1\\ 2\\ 3}=e

ba={1\\ 2\\ 3\\choose 3\\ 1\\ 2}{1\\ 2\\ 3\\choose 2\\ 3\\ 1}={1\\ 2\\ 3\\choose 1\\ 2\\ 3}=e

The other $3\\operatorname{x}3$ blocks are also similar. Now continuing with the upper left 3 x 3 block, we go through the table again using the fact that there can be no repetition in any row or column.

ar={1\\ 2\\ 3\\choose 2\\ 3\\ 1}{1\\ 2\\ 3\\choose 2\\ 1\\ 3}={1\\ 2\\ 3\\choose 3\\ 2\\ 1}=s

as={1\\ 2\\ 3\\choose 2\\ 3\\ 1}{1\\ 2\\ 3\\choose 3\\ 2\\ 1}={1\\ 2\\ 3\\choose 1\\ 3\\ 2}=t

Similarly, we complete the final blocks of the table.

ra={1\\ 2\\ 3\\choose 2\\ 1\\ 3}{1\\ 2\\ 3\\choose 2\\ 3\\ 1}={1\\ 2\\ 3\\choose 1\\ 3\\ 2}=t

rb={1\\ 2\\ 3\\choose 2\\ 1\\ 3}{1\\ 2\\ 3\\choose 3\\ 1\\ 2}={1\\ 2\\ 3\\choose 3\\ 2\\ 1}=s

sa={1\\ 2\\ 3\\choose 3\\ 2\\ 1}{1\\ 2\\ 3\\choose 2\\ 3\\ 1}={1\\ 2\\ 3\\choose 2\\ 1\\ 3}=r

sr={1\\ 2\\ 3\\choose 3\\ 2\\ 1}{1\\ 2\\ 3\\choose 2\\ 1\\ 3}={1\\ 2\\ 3\\choose 2\\ 3\\ 1}=a

Finally, we fill in the table using the calculated values above.

$\\begin{array}[]{|c||c|c|c|c|c|c|}\\hline&e&a&b&r&s&t\\\\\\hline\\hline e&e&a&b&r&s&t\\\\\\hline a&a&b&e&s&t&r\\\\\\hline b&b&e&a&t&r&s\\\\\\hline r&r&t&s&e&b&a\\\\\\hline s&s&r&t&a&e&b\\\\\\hline t&t&s&r&b&a&e\\\\\\hline\\end{array}$