proof of Cauchy’s theorem in abelian case

Suppose $G$ is abelian and the order of $G$ is $h$. Let $g_1$, $g_2,\mathrm{\dots },g_h$ be the elements of $G$, and for $i=1,\mathrm{\dots },h$, let $a_i$ be the order of $g_i$.

Consider the direct sum

$$H=\underset{i=1}{\overset{h}{\oplus }}\mathbb{Z}/a_i\mathbb{Z}.$$

The order of $H$ is obviously $a_1{a}_2\mathrm{\cdots }{a}_h$. We can define a group homomorphism $\theta$ from $H$ to $G$ by

$$(x_1,\mathrm{\dots },x_h)\mapsto g_1^{x_1}\mathrm{\cdots }{g}_h^{x_h}.$$

$\theta$ 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
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