proof of Cauchy’s theorem in abelian case

Suppose G is abelianMathworldPlanetmath and the order of G is h. Let g1, g2,,gh be the elements of G, and for i=1,,h, let ai be the order of gi.

Consider the direct sumMathworldPlanetmath


The order of H is obviously a1a2ah. We can define a group homomorphismMathworldPlanetmath θ from H to G by


θ is certainly surjectivePlanetmathPlanetmath. So |H|=|G||ker(θ)|. Since p is a prime factorMathworldPlanetmath of G, p divides —H—, and therefore must divide one of the ai’s, say a1. Then g1a1/p is an element of order p.

Title proof of Cauchy’s theorem in abelian case
Canonical name ProofOfCauchysTheoremInAbelianCase
Date of creation 2013-03-22 14:30:28
Last modified on 2013-03-22 14:30:28
Owner kshum (5987)
Last modified by kshum (5987)
Numerical id 7
Author kshum (5987)
Entry type Proof
Classification msc 20D99
Classification msc 20E07