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

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?

  • $\begingroup$ I think $(0,1)+<(1,2)>$=$\{(0,1), (1,3)\}$. $\endgroup$ – Thomas Rasberry Apr 26 '17 at 4:08
  • $\begingroup$ and $(1,0)+\langle(1,2)\rangle=\{(1,0),(0,2)\}=(0,2)+\langle(1,2)\rangle $\endgroup$ – Robert Chamberlain Apr 26 '17 at 6:10

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.


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