# Math Help - Question about finite cyclic groups

1. ## Question about finite cyclic groups

Is it true that for every finite cyclic group <G,*> with generator a each of the unique elements of the group may be presented as $a^n$ where n is non-negative integer.

My proof: If G is finite, there is some positive integer m for which $a^m=e$. Then $a^{m+1}=a^1$, $a^{m+2}=a^2$ etc...
Moreover $a^m=a^{-m}=e$. Then $a^m*a^1=a^{-m}*a^1$. Therefore $a^{m+1}=a^{-m+1}=a^1$.
If $a^{m+n}=a^{-m+n}$ for some n<m then $a^{m+n+1}=a^{-m+n+1}$ too.

This must cover all inverses of the elements from $a^1$ to $a^m$.

2. ## Re: Question about finite cyclic groups

well, what you want to do, is show that $a^{-1}$ can be expressed as a positive power of a. and you have almost done done this, which power of a is it?

also, you need to flesh out the existence of m a bit more. all we're given to work with, is that <a> is finite.

i would argue thusly: there are infinitely many positive powers of a, but since G = <a> is finite, for some r ≠ s we must have $a^r = a^s$.

without loss of generality, we may assume that r > s. then $a^ra^{-s} = a^{r-s} = e$. this shows that for SOME positive integer n, namely n= r-s,

we have $a^n = e$. we can then let m be the smallest positive integer such that $a^m = e$ (because we know there is at least one).

this gives us a set of DISTINCT positive powers of a $\{a,a^2,\dots,a^m = e\}$. your proof that ANY positive power of a is in this set consists mainly of the word: "etc."

you can do better. write an arbitrary positive number n, as n = qm + r, where 0 ≤ r < m (why can we do this?). show that $a^n = a^r$.

if $a^{-1}$ is in this set, then so is $a^{-k} = (a^{-1})^k$ (why?)

3. ## Re: Question about finite cyclic groups

This forum seems to be infected with malware. I'll stop posting for a while until this is sorted out. I understand the proof of existence of m. n = qm + r comes from division algorithm. Thanks for the advice!