Group theory: Let $H=\{0,\pm 3, \pm 6, \pm9,\ldots\}$ Find all the left cosets of $H$ in $\Bbb Z$. I am having trouble understanding the following homework question,
Let $H=\{0,\pm 3, \pm6, \pm9,\ldots\}$ Find all the left cosets of $H$ in $\Bbb Z$.
I know the answer is $H$, $1+H$, and $2+H$ but I am having difficulty understanding why.
Thank you!

Observe that $H=3\Bbb Z$. The fact that $H,1+H,2+H$ are the only cosets is a restatement of the fact any integer $n$ is of the form $3k+b$ with $b=0,1,2$.
