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I am referring to this proof I have a few questions:

  1. Is $\mathbb Z$ here, the centralizer of $\mathbb G$ ?
  2. How did they arrive at this conclusion?

The order of $\mathbb H$ is obviously $a_ia_2...a_h$

Thanks for your help Soham

Note, the relevant part of the proof for this question is that $G$ is a finite group with $h$ elements $g_1, g_2 \dots g_h$ having orders $a_1, a_2, \dots a_h$ respectively. We define a group $H$ of order $a_1a_2\dots a_h$ as follows:

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

and proceed to use a homomorphism from $H$ to $G$ to prove Cauchy's theorem.

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(1) No. It is the integers group. –  DonAntonio Sep 4 '12 at 10:00
    
@DonAntonio Integers group??? :-o Hmm never saw such strange things. Can we bring a specificity in a proof. Isnt proof supposed to be the most of generalized "proof" ? –  Soham Sep 4 '12 at 10:05
    
I think you didn't understand that part of the proof: he uses residue groups to do the proof, and there's where the integers get in the game. BTW, I recommend you MacKay's proof of Cauchy's Theorem: way more beautiful, elegant...and doesn't need to separate in abelian and non-abelian cases. Try here proofwiki.org/wiki/Cauchy's_Group_Theorem –  DonAntonio Sep 4 '12 at 10:22
    
@DonAntonio May sound knaive, but for a moment I just wished if I could study algebra under you :) –  Soham Sep 4 '12 at 14:58

1 Answer 1

up vote 3 down vote accepted

$\mathbb Z/a_i\mathbb Z$ in the sum is the additive group of integers taken $\operatorname{mod} a_i$. There are $a_i$ elements of the group. When you select an element of the direct sum you take a contribution from each component, so to get the order of the group you multiply the $a_i$ together.

$H$ is a manufactured group, which is constructed specially for the purpose of the proof. It is built up using known groups, one for each element of the unknown finite group $G$. The existence of a homomorphism from the known group $H$ to the unknown group $G$ enables us to discover information about $G$ using what we know about $H$.

This is a proof strategy replicated many times, and very useful to know.

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Much thanks, it clearly looks a very intelligent way of doing things. Thank you –  Soham Sep 4 '12 at 10:21

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