Let $ \scriptstyle p $ be a prime. Show that in a cyclic group of order $ \scriptstyle p^n-1 $, every element is a $ \scriptstyle p $th power (that is, every element can be written in the form $ \scriptstyle a^p $ for some $ \scriptstyle a $).
Consider a cyclic group $ \scriptstyle G $ of order $ \scriptstyle p^n-1 $, where $ \scriptstyle p $ is prime. Because $ \scriptstyle G $ is cyclic, there exists at least one generator $ \scriptstyle a\mid a\in G $. Thus, $ \scriptstyle G=\langle a\rangle $. Since $ \scriptstyle gcd(p^n-1,p)=1 $, by Corollary 2 of Theorem 4.2, $ \scriptstyle G=\langle a^p\rangle $. This means that every element of $ \scriptstyle G $ is contained in the list $ \scriptstyle a^{p^1},a^{p^2},\ldots,a^{p^{p^n-1}} $. But then, this list may be equivalently written as $ \scriptstyle (a^1)^p,(a^2)^p,\ldots,(a^{p^n-1})^p $. It is then clear that every element of $ \scriptstyle G $ is a $ \scriptstyle p $th power of some $ \scriptstyle a $.
- --Nick Rupley 04:56, 4 February 2009 (UTC)
I came up with a similar solution. I didn't understand it so well untill I read this though.
The another probable solution could be (since I am not sure about it's legal operations) the following: Since G is cyclic, we have some generator $ \scriptstyle a $ that has this property. $ \scriptstyle a^{p^n-1}=1 $. In other words, $ \scriptstyle a^{p^n}=a $. Well, then that means that now any element is represented as a multiplication of $ \scriptstyle a^{p^n} $ which shows that any element is represented by the power of p since $ \scriptstyle a $ is a generator.
The only problem that I have with this is how much of a legal operation I have done in $ \scriptstyle a^{p^n}=a $ by multiplying both sides with $ \scriptstyle a $.
