> [!NOTE] Theorem (Powers of a group element only generate other group elements) > Let $G$ be a [[Groups|group]]. Let $a\in G.$ Then the [[Integer Power of Group Element|powers]] of $a$ are all elements of $G$: that is, in [[Multiplicative Notation|multiplicative notation]], $a^{n}\in G$ for all $n\in\mathbb{Z}.$ *Proof*. If $n=0,$ $a^{0}=1$ and certainly $1\in G.$ Suppose that $a^{k}\in G$ for some $k\in\mathbb{N}.$ Then $a^{k+1} =a^{k}a \in G$ since both $a^{k}$ and $a$ are in $G$ which is closed under the binary operation. It follows by induction that $a^{n}\in G$ for all $n\in \mathbb{N}.$ On the other hand, if $n$ is a negative integer then $a^{n}=(a^{-n})^{-1}.$ But we know that $a^{-n}\in G$ since $-n$ is positive and so $(a^{-n})^{-1} \in G$ since every group element has an inverse in $G.$ # Applications See [[Generated Subgroup]].