The question I'm working on is:
Let $G=\mathbb{Z}_4\times\mathbb{Z}_3\times\mathbb{Z}_2$ and consider the subgroup $H=\langle\left(0,1,1\right)\rangle$ of G. Find all cosets of H.
So I know that in G, the generator $(0,1,1)$ generates the set $H= \{(0,0,0),(0,0,1),(0,1,0),(0,2,0),(0,1,1),(0,2,1)\}$. and that to find all of the left cosets of H, represented by $x$ (an element of G) is the set \begin{align} xH=\{xh | h\in H\} \end{align}
But I'm a little confused about how to handle this problem with the element H, and not just simply when say $H=\langle 9 \rangle$
Pick an element $x$ in $G$, not in $H$. Then $H+x=\{h+x|h\in H\}$ gives you a new coset. If there are still elements in $G$ that have not appeared in any coset, pick one and make its coset.
To be specific, we might choose $(1,0,0)\notin H$ and form the coset
$$H+(1,0,0)=\{(1,0,0),(1,0,1),(1,1,0),(1,2,0),(1,1,1),(1,2,1)\}$$
There are still elements of $G$ that don't appear in $H$ or $H+(1,0,0)$, so you pick one of these elements and start a new coset.
Note: since each coset has $6$ elements, and $G$ has $24$ elements, there will be a total of $4$ cosets (including $H$).