proof that every group of prime order is cyclic

Let p be a prime and G be a group such that |G|=p. Then G contains more than one element. Let gG such that geG. Then g contains more than one element. Since gG, by Lagrange’s theorem, |g| divides p. Since |g|>1 and |g| divides a prime, |g|=p=|G|. Hence, g=G. It follows that G is cyclic.

Title proof that every group of prime order is cyclic
Canonical name ProofThatEveryGroupOfPrimeOrderIsCyclic
Date of creation 2013-03-22 13:30:55
Last modified on 2013-03-22 13:30:55
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 7
Author Wkbj79 (1863)
Entry type Proof
Classification msc 20D99
Related topic ProofThatGInGImpliesThatLangleGRangleLeG