# 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 $ab$ and $ba$ 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.

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

Title symmetric group on three letters SymmetricGroupOnThreeLetters 2013-03-22 15:52:24 2013-03-22 15:52:24 Wkbj79 (1863) Wkbj79 (1863) 13 Wkbj79 (1863) Example msc 20B30