Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

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: $\,x+H=y+H\Longleftrightarrow -x+y\in H\,$

share|improve this answer
    
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  
$(1,2)+(0,2)=(1,0)$ ;) –  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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.