# 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 $n$.

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

and
$g = \left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}&{{b_3}}& \cdots &{{b_n}} \\ {{c_1}}&{{c_2}}&{{c_3}}& \cdots &{{c_n}} \end{array}} \right)$ 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 $fg \in {P_n}$, $\forall f,g \in {P_n}$.