# Finite cyclic groups

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?

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Well, if $g^n=1$, then which of $1,g,g^2,...,g^{n-1}$ would $g^{-1}$ be? – user5137 Feb 7 '12 at 2:04
Ohhhh, is it because the operations are multiplication and addtion? – Shannon Feb 7 '12 at 2:17
Not quite...we're in a group, so there's really only one operation that we care about (and which you've denoted multiplicatively in your question). – user5137 Feb 7 '12 at 2:24

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 $$G=\{g^{k_1},g^{k_2},\dots,g^{k_n}\}\;.$$
 Great! For the exercise 2, can you tell me which g^k_i=1? – Shannon Feb 7 '12 at 2:27 @Shannon: You could choose it to be any of them. As a hint for the exercise, think about the exponents modulo $n$. – Brian M. Scott Feb 7 '12 at 2:30