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 multiplication 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.

$$e={ 1\ 2\ 3 \choose 1\ 2\ 3} \hspace{20mm} r ={ 1\ 2\ 3 \choose 2\ 1\ 3}$$ $$a={ 1\ 2\ 3 \choose 2\ 3\ 1} \hspace{20mm} s ={ 1\ 2\ 3 \choose 3\ 2\ 1}$$ $$b={ 1\ 2\ 3 \choose 3\ 1\ 2} \hspace{20mm} t ={ 1\ 2\ 3 \choose 1\ 3\ 2}$$

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.