Take the 2-minute tour ×
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.

In $\mathbb{Z}/16$ write down all the cosets of the subgroup H={[0],[4],[8],[12]}.

This is what I have:

o+[0]

1+[0]

2+[0]

3+[0]

0+[4]

1+[4]

2+[4]

3+[4]

0+[8]

1+[8]

2+[8]

3+[8]

0+[12]

1+[12]

2+[12]

3+[12]

Is this right or am I missing some cosets?

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

I think that you are missing something. A coset of $H$ in $G = \mathbb{Z}/16\mathbb{Z}$ would be something like $[1] + H$. Note for example that $[1]+ H = [9]+H$ because $[9] - [1] = [8]\in H$.

Edit 1: Note for example that the number of cosets will equal:

$$ \lvert G / H\lvert = \lvert G\lvert / \lvert H\lvert = 16 / 4 = 4. $$

Edit 2: So the the cosets are: $$ \begin{align} &[0] + H = [4] + H = [8] + H = [12] + H \\ &[1] + H = \dots \\ &[2] + H = \dots \end{align} $$ Can you find the last one?

share|improve this answer
    
So I should have only written [0]+H,[1]+H,[2]+H,[3]+H. Is that what you mean? –  user39794 Oct 23 '12 at 15:14
1  
@AllisonCameron: Yes. What you have written in your question aren't cosets. –  Thomas Oct 23 '12 at 15:20
    
@AllisonCameron: I edited my answer to give a bit more help. –  Thomas Oct 23 '12 at 15:22
    
Thank you for the help! :) –  user39794 Oct 23 '12 at 15:31
add comment

Your Answer

 
discard

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