# Group of Permutations

The set $${P_n}$$ of all permutations on $$n$$ symbols is a finite group of order $$n{!}$$ with respect to the 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 the multiplication of permutation on $${P_n}$$ satisfies 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 is 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 the 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 permutations of degree $$n$$ defined on a finite set forms a finite group of order $$n!$$ with respect to the composite of permutations 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 permutations of degree 2, i.e. the group of all permutations defined on a set of two elements $$\left( {{a_1},{a_2}} \right)$$, then

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

Now

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

and

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

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