# Left cosets

• Jan 11th 2010, 09:22 AM
Amer
Left cosets
find all left coset taking the subgroup generated by $<(1,2)>$

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

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

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

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

$h+(0,0) = h$

$h+(1,0)$

$h+(0,1)$

$h+(1,1)$

how the book take these elements can you help
is there a way to chose these elements (Thinking)
• Jan 11th 2010, 09:33 AM
Drexel28
Quote:

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

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

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

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

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

$h+(0,0) = h$

$h+(1,0)$

$h+(0,1)$

$h+(1,1)$

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

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

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.
• Jan 11th 2010, 09:41 AM
Amer
Quote:

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

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 :)
• Jan 11th 2010, 09:43 AM
Drexel28
Quote:

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.