> [!NOTE] Definition (Integer power of group element) > Let $(G,\circ)$ be a [[Groups|group]] and $a\in g.$ Let $n \in \mathbb{Z}.$ The $n$th power of $a$ is defined [[Recursive Function|recursively]] using [[Multiplicative Notation|multiplicative notation]] as $a^{n} = \begin{cases} 1 & n = 0, \\ a^{n-1} \circ a & n>0, \\ (a^{-n})^{-1} & \text{otherwise.} \end{cases}$or using additive notation $na = \begin{cases} 0 &n=0,\\ ((n-1)a) \circ a & n>0, \\ -((-n)a) & \text{otherwise.} \end{cases}$ # Properties By [[Powers of a group element only generate other group elements]], $a^{n}\in G$ for all $n\in \mathbb{Z}$ and $a\in G.$ By [[Inverse of Power of Group Element]], for all $a\in G,$ $n\in \mathbb{Z},$ $(a^{-1})^{n}=a^{-n}=(a^{n})^{-1}.$ By [[Power of Power of Group Element]], for all $a\in G,$ $m,n\in \mathbb{Z},$ $(a^{m})^{n}=a^{mn}.$ By [[Product of Powers of a Group Element]], for all $a\in G,$ $m,n\in \mathbb{Z},$ $a^{m}a^{n}=a^{m+n}.$ By [[Power of Product of Abelian Group Elements]], $(ab)^{n}=a^{n}b^{n}$ for all $a,b\in G$ where $G$ is abelian. By [[Criteria for Equality of Powers of Group Element]], Let $a\in G$ with infinite order then $a^{i}=a^{j}$ iff $i=j$ otherwise $a^{i}=a^{j}$ iff $n\mid(i-j)$ where $n$ is the order of $a.$ # Applications **Order**: The [[Order of Group Element|order of a group element]] $a$ is defined as the smallest natural number $n$ that satisfies $a^{n}=1.$ If no such positive integer exists $a$ is said to have infinite order. By [[Order of power of group element]], $|a^{k}|= \frac{|a|}{\gcd(k,|a|)}.$ **Generated subgroup**: The [[Generated Subgroup|subgroup generated]] by a group element $a$ is the set of all integer powers of $a.$ The order of this group is the same the order of $a.$