# 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 $$\forall m,n \in I$$