1. ## Left cosets

find all left coset taking the subgroup generated by $\displaystyle <(1,2)>$

the group is $\displaystyle \mathbb{Z}_2\times \mathbb{Z}_4$

we have the subgroup $\displaystyle h={(0,0),(1,2)}$

$\displaystyle \mathbb{Z}_2\times \mathbb{Z}_4 =[(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)]$

I take all elements not in h and add them to h to get the left coset but I have repeated elements so how I can chose elements such that I do not have repeated ones

the book said

$\displaystyle h+(0,0) = h$

$\displaystyle h+(1,0)$

$\displaystyle h+(0,1)$

$\displaystyle h+(1,1)$

how the book take these elements can you help
is there a way to chose these elements

2. Originally Posted by Amer
find all left coset taking the subgroup generated by $\displaystyle <(1,2)>$

the group is $\displaystyle \mathbb{Z}_2\times \mathbb{Z}_4$

we have the subgroup $\displaystyle h={(0,0),(1,2)}$

$\displaystyle \mathbb{Z}_2\times \mathbb{Z}_4 =[(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)]$

I take all elements not in h and add them to h to get the left coset but I have repeated elements so how I can chose elements such that I do not have repeated ones

the book said

$\displaystyle h+(0,0) = h$

$\displaystyle h+(1,0)$

$\displaystyle h+(0,1)$

$\displaystyle h+(1,1)$

how the book take these elements can you help
is there a way to chose these elements
Do you know Lagrange's theorem? Since $\displaystyle \left|\mathbb{Z}_2\oplus \mathbb{Z}_4\right|=8$ and $\displaystyle \text{ord }\left(1,2\right)=2\implies\left|\left \langle \left(1,2\right)\right \rangle\right|=2$ we mus only find $\displaystyle \frac{8}{2}=4$ subgroups. And, since the subgroup itself is always a coset we must only find three more subgroup. So how about the four cosets $\displaystyle \left\langle \left(1,2\right)\right \rangle=H,\left(1,1\right)+H,\left(0,2\right)+H,\l eft(0,3\right)+H$?

There is no diehard way, that I know to pick the elements. In this case, and some others, you can eye-ball the subgroup to see what will give you distinct cosets.

3. Originally Posted by Drexel28
Do you know Lagrange's theorem? Since $\displaystyle \left|\mathbb{Z}_2\oplus \mathbb{Z}_4\right|=8$ and $\displaystyle \text{ord }\left(1,2\right)=4\implies\left|\left \langle \left(1,2\right)\right \rangle\right|=2$ we mus only find $\displaystyle \frac{8}{2}=4$ subgroups. And, since the subgroup itself is always a coset we must only find one more subgroup. So how about the four cosets $\displaystyle \left\langle \left(1,2\right)\right \rangle=H,\left(1,1\right)+H,\left(0,2\right)+H,\l eft(0,3\right)+H$?

There is no diehard way, that I know to pick the elements. In this case, and some others, you can eye-ball the subgroup to see what will give you distinct cosets.
I know it, and I know I have 4 subgroups in the left coset

it is clear now thank you very much

4. Originally Posted by Amer
I know it, and I know I have 4 subgroups in the left coset

it is clear now thank you very much
I made a typo! I'm sure you paid no mind to it, but I though I should point it out.