# 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.