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)

Copied from problem 61. I was not smart enough to realize 61 (6th ed) was the same as 63 (7th ed), until I clicked 61 on accident. Nick's solution belongs here too! --Bcaulkin 00:37, 5 February 2009 (UTC)

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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

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