proof of Cauchy’s theorem in abelian case

Suppose $G$ is abelian and the order of $G$ is $h$. Let $g_{1}$, $g_{2},\ldots,g_{h}$ be the elements of $G$, and for $i=1,\ldots,h$, let $a_{i}$ be the order of $g_{i}$.

Consider the direct sum

 $H=\bigoplus_{i=1}^{h}\mathbb{Z}/a_{i}\mathbb{Z}.$

The order of $H$ is obviously $a_{1}a_{2}\cdots a_{h}$. We can define a group homomorphism $\theta$ from $H$ to $G$ by

 $(x_{1},\ldots,x_{h})\mapsto g_{1}^{x_{1}}\cdots g_{h}^{x_{h}}.$

$\theta$ is certainly surjective. So $|H|=|G|\cdot|\ker(\theta)|$. Since $p$ is a prime factor of $G$, $p$ divides —H—, and therefore must divide one of the $a_{i}$’s, say $a_{1}$. Then $g_{1}^{a_{1}/p}$ is an element of order $p$.

Title proof of Cauchy’s theorem in abelian case ProofOfCauchysTheoremInAbelianCase 2013-03-22 14:30:28 2013-03-22 14:30:28 kshum (5987) kshum (5987) 7 kshum (5987) Proof msc 20D99 msc 20E07