Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
(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'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
up vote 4 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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.