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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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1 Answer 1

up vote 3 down vote accepted

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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I got confused because the two cosets don't contain the same elements. $(1,0)$ is not in $(1,2)+H$, but is in $(1,1)+H$. – rick Oct 9 '12 at 3:53
$(1,2)+(0,2)=(1,0)$ ;) – N. S. Oct 9 '12 at 3:57
@rick , N.S. already answered you. It seems like you forgot you're workimg modulo 2 in the left coordinate and modulo 4 in the right one. – DonAntonio Oct 9 '12 at 4:04

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