# 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 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:

$\forall a \in G$ 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:

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