Group of Permutations

The set {P_n} of all permutations on n symbols is a finite group of order n{!} with respect to composite of mappings as the operation. For n \leqslant 2, this group is abelian and for n > 2 it is always non-abelian.

Let S = \left\{ {{a_1},{a_2},{a_3}, \ldots ,{a_n}}  \right\} be a finite set having n distinct elements. Thus there are n{!} permutations possible on S. If {P_r} denotes the set of all permutations of degree n then multiplication of permutation on {P_n}satisfy the following axioms.

Closure Axiom: Let f,g \in {P_n} then each of them is one-one mapping of S onto itself and therefore their composite mapping \left(  {g \circ f} \right) is a one-one mapping of S onto itself. Thus \left( {g \circ f} \right) is a permutation of degree n on S, i.e.

 f,g \in {P_n} \Rightarrow fg \in {P_n}

This shows that {P_n} is closed under multiplication.

Associative Axiom: Since the product of two permutations on a set S is nothing but the product of two one-one onto mappings on S and the product of mapping being associative, the product of permutations also obeys the associative law. Hence

 f,g,h \in {P_n} \Rightarrow \left( {fg} \right)h = f\left(  {gh} \right)


Identity Axiom: Identity permutation I \in {P_n} is identity of multiplication in {P_n} because

If = fI = f\,\,\,\forall f \in {P_n}

Inverse Axiom: Let f \in {P_n} then f is one-one mapping hence it is investible. Hence {f^{ - 1}}, the inverse mapping of f is also one-one and onto. Consequently, {f^{ - 1}} is also a permutation in {P_n}.

 {f^{ - 1}}f = f{f^{ - 1}} = I

Thus the symmetric set {P_n} of all permutation of degree n defined on a finite set forms a finite group of order \(n\) with respect to the composite of permutation as the composition.

Commutative Axiom: If we consider the symmetric group \left(  {{P_1},0} \right) of permutations of degree 1 with respect to permutation product 0, then it consists of a single permutation namely the identity permutation I. Since I0I = I, \left(  {{P_1},0} \right) is an Abelian group. If we consider the symmetric group \left( {{P_2},0} \right) of all permutation of degree 2, i.e. the group of all permutations defined on a set of two elements \left( {{a_1},{a_2}}  \right), then

 {P_2} = \left\{ {\left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}  \\ {{a_1}}&{{a_2}} \end{array}} \right),\left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}  \\ {{a_2}}&{{a_1}} \end{array}} \right)} \right\}


 \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}} \\ {{a_1}}&{{a_2}}  \end{array}} \right)0\left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}} \\ {{a_2}}&{{a_1}}  \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}} \\ {{a_2}}&{{a_1}}  \end{array}} \right)


\left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}} \\ {{a_2}}&{{a_1}}  \end{array}} \right)0\left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}} \\ {{a_1}}&{{a_1}}  \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}} \\ {{a_2}}&{{a_1}}  \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{a_2}}&{{a_1}} \\ {{a_2}}&{{a_1}}  \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}} \\ {{a_2}}&{{a_1}}  \end{array}} \right)

Therefore operation having commutative \left( {{P_2},0} \right) is Abelian group of order 2. But when n >  2 then permutation product is not necessarily commutative. Hence \left( {{P_n},0} \right) then is not necessarily an Abelian group.