Given the group $G=(\mathbb{Z}_{12},+)$ and a subgroup $H=\langle[4]\rangle$, list the left cosets of $H$.
All that I understand about (left) cosets is that $x\sim y\iff x=yh$ where $h\in H$. I don't know how to 'calculate' them though.
The answers that were given for this question are as follows:
- $[1]+H=\{[1],[5],[9]\}$
- $[2]+H=\{[2],[6],[10]\}$
- $[3]+H=\{[3],[7],[11]\}$
I can see from context how these sets were calculated ($[1]+[0]=[1]$, and so on).. but how were $[1], [2], [3]$ chosen to be added with the elements of $H$?
Best Answer
1) We know that $\mathbb{Z}_{12}=\{0,1,2,3,\cdots,11\}$ and $H=\{0,4,8\}$ and we are working with abelian finite groups.
2) The group is finite of order $12$ so we have $\big|\frac{\mathbb{Z}_{12}}{H}\big|=\frac{12}{3}=4$.
3) The quotient group $\frac{\mathbb{Z}_{12}}{H}$ is $\{gH|g\in\mathbb{Z}_{12}\}=\{g+H|g\in\mathbb{Z}_{12}\}$ and know that $$g+H=\{g+h|h\in H\}$$
4) We learned that if suddenly $g\in H$ then $g+H=H$ which is $e_{\frac{\mathbb{Z}_{12}}{H}}$.
5) So we have, finally, $$\frac{\mathbb{Z}_{12}}{H}=\{1+H,2+H,3+H, 5+H, 6+H, 7+H, 9+H,10+H, 11+H\}$$
6) Now, lets to do some handy calculations. What is $9+H$? It is $$\{9+0,9+4,9+8\}=\{9,1,5\}=\{1+8,1+0,1+4\}=1+H$$
7) Do the same way to find your final answer.