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?

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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?

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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
@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