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I came up with a similar solution. I didn't understand it so well untill I read this though. | I came up with a similar solution. I didn't understand it so well untill I read this though. | ||
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| + | 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 a that has this property. | ||
| + | <math>\scriptstyle a^{p^n-1}=1</math>. | ||
Revision as of 09:52, 5 February 2009
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 a that has this property. $ \scriptstyle a^{p^n-1}=1 $.
