left cosets of Z2 x Z4

So I am having a problem. The question is:

Find the left coset of <(1,0)> in Z₂ x Z₄.

Since there are 8 elements in Z₂ x Z₄₂ and only 2 in (1,0), there should be 4 left cosets, but I can only find one.. (1,0). What am I missing?

Well, I'm not sure what you're missing without seeing your working out but firstly are you correctly finding the subgroup?

<(1,0> = {(1,0), (0,0)}

Then, the process is:

Take a particular element in Z_{2} x Z_{4} and add it to each of the elements in <(1,0)>. Do this for each element in Z_{2} x Z_{4}. This will give you 8 cosets, but as you say you should find there are only 4 DISJOINT cosets each with two elements.

Hi cloeannx3,

You're correct that there should be 4 left/right cosets - nice job! I'll work out two cosets and leave the other two for you.

Let's start by writing out the following:

<(1,0)> = $\displaystyle \{(0,0), (1,0)\}$

$\displaystyle \mathbb{Z}_{2}\times\mathbb{Z}_{4}=\{(0,0), (0,1), (0,2), (0,3), (1,0), (1,1), (1,2), (1,3)\}.$

To determine what the (left) cosets are we go down the list of elements in $\displaystyle \mathbb{Z}_{2}\times \mathbb{Z}_{4}$ making cosets of <(1,0)>. The first two cosets are

1) $\displaystyle (0,0) + <(1,0)> = \{(0,0), (1,0)\}=<(1,0)>$

2) $\displaystyle (0,1) + <(1,0)> = \{(0,1), (1,1)\}.$ Notice that this (left) coset is exactly the same as the coset $\displaystyle (1,1) + <(1,0)>.$

Notice that the sets two cosets we have found in 1 and 2 are disjoint.

Does this help get things on the right track? Let me know if anything is unclear. Good luck!

That make so much more sense! Thank you so much!!