# Integral Powers of an Element of a Group

Suppose $G$ is a group and the composition has been denoted by multiplicatively, let $a \in G$. Then by closure property $a,aa,aaa,aaaa$ etc., are all elements of $G$. Since the composition in $G$ obeys general associative law, therefore $aaa…a$ to $n$ factors is dependent of the manner in which the factors may be grouped.

If $n$ is a positive integer, we define ${a^n} = aaa…a$ factors to $n$ factors. Obviously ${a^n} \in G$. If $n$ is the identity element of the group $G$, then we define ${a^0} = e$.

If $n$ is a positive integer then $– n$ is a negative integer. Now we define ${a^{ – n}} = {\left( {{a^n}} \right)^{ – 1}}$ where ${\left( {{a^n}} \right)^{ – 1}}$ is the inverse of ${a_n}$in $G$. Thus, ${a^{ – n}} \in G$. Thus we have defined ${a^n}$ for all integral values of $n$ positive, zero or negative.

Integral Multiples of an Element of a Group

If in a group $G$ the composition has been denoted additively, then in place of using the word integral powers of an element of a group we use the word integral multiples of an element of a group. The difference is only of notation, otherwise the meaning is the same. Thus in this case if $n$ is a positive integer we write $na$ in place of ${a^n}$ and we define $na = a + a + \cdots + a$ up to $n$ terms.

In place of ${a^0}$ we write ${O_a}$. Thus we define ${O_a} = e$ where $e$ is the identity of $G$.

If $n$ is a positive integer, then in place of ${a^{ – n}}$ we write $\left( { – n} \right)a$.

Thus, we define $\left( { – n} \right)a = – \left( {na} \right)$ where $– \left( {na} \right)$ denotes the inverse of $na$ in $G$.

In multiplicative notation the following laws of indices can be easily proved:
${a^m}{a^n} = {a^{m + n}}$
${\left( {{a^m}} \right)^n} = {a^{mn}}$

$\forall a \in G$ and $\forall m,n \in I$ where $I$ is the set of integers.

In additive notation the following laws of multiples can be easily proved:
$ma + na = \left( {m + n} \right)a$
$n\left( {ma} \right) = \left( {mn} \right)a$

$\forall a \in G$ and $\forall m,n \in I$