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'proof of Cauchy's theorem in abelian case'
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| Title of object: |
proof of Cauchy's theorem in abelian case |
| Canonical Name: |
ProofOfCauchysTheoremInAbelianCase |
| Type: |
Proof |
| Created on: |
2004-07-29 15:22:13 |
| Modified on: |
2004-07-29 15:22:13 |
| Classification: |
msc:20D99, msc:20E07 |
Preamble:
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Content:
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 product
H = \oplus_{i=1}^h \mathbb{Z}/a_i\mathbb{Z}.
The order of $H$ is obviously $a_1a_2\cdots a_h$. 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}.
The homomorphism 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 $p$. |
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