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:

{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

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