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 $$fg \in {P_n}$$, $$\forall f,g \in {P_n}$$.