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?
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?
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?