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If you have a group $G$ and $H \leq G$, the cosets of $H$ should partition $G$.

Suppose $G=\mathbb{Z}_2\times\mathbb{Z}_4$ and $H=\langle (0,1)\rangle = \{(0,0), (0,1),(0,2),(0,3)\}$. Then both $(1,1)+H$ and $(1,2)+H$ contain $(1,3)$. What am I doing that's dumb?

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Nothing. Indeed

$(1,2)+H=(1,1)+H\Longleftrightarrow (1,2)-(1,1)\in H$

Remember: $\,xH=yH\Longleftrightarrow x^{-1}y\in H\,$ , or additively: $\,x+H=y+H\Longleftrightarrow -x+y\in H\,$

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    @rick , N.S. already a$n$swered you. It seems like you forgot you're workimg modulo 2 i$n$ the left coordi$n$ate a$n$d modulo 4 in the right one.2012-10-09