Finite cyclic groups

Question

According to the theorem that a finite cyclic subgroup $G=\langle g \rangle$, there exsits a smallest positive integer $n$ such that $g^n=1$, and we have $G=\{1,g,g^2,...,g^{n-1}\}$ where $1,g,...,g^{n-1}$ are all distinct. I am just wondering why those powers couldn't be negtive? Is it because $G$ is finite?

Answer 1

They could: $G$ is also equal to $\{1,g^{-1},g^{-2},\dots,g^{1-n}\}$ and to $\{g^n,g^{n+1},\dots,g^{2n-1}\}$, among many other sets of powers of $g$. It’s pretty easy to see that $G=\{g^n,g^{n+1},\dots,g^{2n-1}\}$: after all, $g^{k+n}=g^k\cdot g^n=g^k\cdot 1=g^k$, so $g^n=1,g^{n+1}=g,\dots,g^{2n-1}=g^{n-1}$. You have to work a little harder to match up $\{1,g,g^2,\dots,g^{n-1}\}$ with $\{1,g^{-1},g^{-2},\dots,g^{1-n}\}$, but not much: $g\cdot g^{n-1}=1$, so $g^{-1}=g^{n-1}$, and in general $g^{n-k}=g^n\cdot g^{-k}=1\cdot g^{-k}=g^{-k}$.

Exercise: If $k,k+1,\dots,k+n-1$ are any $n$ consecutive integers, then $$G=\{g^k,g^{k+1},\dots,g^{k+n-1}\}\;.$$

Exercise: Find a set of $n$ exponents, $\{k_1,k_2,\dots,k_n\}$, that are not consecutive integers but still have the property that $$G=\{g^{k_1},g^{k_2},\dots,g^{k_n}\}\;.$$