Product or Composite of Two Permutations

The products or composite of two permutations f and g of degree n denoted by fg is obtained by first carrying out the operation defined by f and then by g.

Let us suppose {P_n} is the set of all permutations of degree n.


 f = \left( {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}& \cdots &{{a_n}} \\ {{b_1}}&{{b_2}}&{{b_3}}& \cdots &{{b_n}} \end{array}} \right)


 g = \left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}&{{b_3}}& \cdots &{{b_n}} \\ {{c_1}}&{{c_2}}&{{c_3}}& \cdots &{{c_n}} \end{array}} \right)

be two elements of {P_n}.

Hence the permutation g has been written in such a way that the first row of g coincides with the second row of f. If the product of the permutations f and g is denoted multiplicatively, i.e., by fg, then by definition

 fg = \left( {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}& \cdots &{{a_n}} \\ {{c_1}}&{{c_2}}&{{c_3}}& \cdots &{{c_n}} \end{array}} \right)

For f replaces {a_1} by{b_1} and then g replaces {b_1} by {c_1} so that fg replaces {a_1} by{c_1}. Similarly fg replaces {a_2} by {c_2}, {a_3} by{c_3}, …,{a_n} by {c_n}.

Obviously, fg is also a permutation of degree n. Thus the product of two permutations of degree n is also a permutation of degree n. Therefore fg \in {P_n}, \forall f,g \in {P_n}.