# Finding cosets of $(\mathbb{Z}_2\times \mathbb{Z}_4)/\langle (1,2)\rangle$

$$(\mathbb{Z}_2\times \mathbb{Z}_4)/\langle (1,2)\rangle$$

In order to find cosets of this, I realized that $\langle (1,2)\rangle = \{(0,0),(1,2)\}$ and thus the order of the quotient is $2\cdot 4/2 = 4$. I know that the cosets are elemnts of $(\mathbb{Z}_2\times \mathbb{Z}_4)$ + the generated $\langle (1,2)\rangle$, so they're like this:

$$(0,0) + \langle (1,2)\rangle = \langle (1,2)\rangle$$ $$(0,1) + \langle (1,2)\rangle = \{(0,1),(0,3)\}$$ $$(0,2) + \langle (1,2)\rangle = \{(0,2),(1,0)\}$$ $$(0,3) + \langle (1,2)\rangle = \{(0,3),(1,1)\}$$ $$(1,0) + \langle (1,2)\rangle = \{(0,0),(0,2)\}$$

all these four are differnt, so they're all the representants of the cosets? Is there a more efficient way to compute this?

Also, see that this group, having $4$ elements, but me isomorph by $\mathbb{Z}_2\times \mathbb{Z}_2$ or $\mathbb{Z}_4$, right? I've seen techniques like finding that an element of the quotient has some order that he element in one of these possibilities doesn't have. How do I know which order to test?

• I think $(0,1)+<(1,2)>$=$\{(0,1), (1,3)\}$. – Thomas Rasberry Apr 26 '17 at 4:08
• and $(1,0)+\langle(1,2)\rangle=\{(1,0),(0,2)\}=(0,2)+\langle(1,2)\rangle – Robert Chamberlain Apr 26 '17 at 6:10 ## 1 Answer The five aren't different, you have some errors in your calculations (see the comments). In general, for a group$G$and subgroup$H$of$G$, the cosets of$H$form a partition of$G$. This means you can find all the cosets by starting with the set$S=\{H\}$then while$\cup_{K\in S}K\ne G$choose$g\in G\setminus\cup_{K\in S}K$and add$gH$to$S\$.

In your case you would obtain the first four cosets and finish.