Coset problem

[Hom]

Find the number of left coset of cyclic group generated by <1,1> of Z/2Z * Z/4Z ?

The answer is 2 according to REA. I don't see why. please help~

[Topoltergeist]

The subgroup generated by (1,1) will include the following four elements:

(1,1)
(0,2)
(1,3)
(0,0)

Another way that you could deduce that this group has 4 elements is that it is the product of a 2-cycle in Z/2Z and a 4-cycle in Z/4Z. The least common multiple of 2 and 4 is 4, hence the subgroup generated has order 4.

That being said, use Lagrange's theorem: for a finite group G with a subgroup H, we have |G| = [G:H] |H|. Let G be the group Z/2Z X Z/4Z and let H be the subgroup generated by (1,1). Following through with the computation we obtain the answer:

| Z/2Z X Z/4Z | = |Z/2Z| * |Z/4Z| = 2*4 = 8
|H| = 4

8 = [G:H] 4
[G:H] = 2

[Hom]

The subgroup generated by (1,1) will include the following four elements:

(1,1)
(0,2)
(1,3)
(0,0)

Another way that you could deduce that this group has 4 elements is that it is the product of a 2-cycle in Z/2Z and a 4-cycle in Z/4Z. The least common multiple of 2 and 4 is 4, hence the subgroup generated has order 4.

That being said, use Lagrange's theorem: for a finite group G with a subgroup H, we have |G| = [G:H] |H|. Let G be the group Z/2Z X Z/4Z and let H be the subgroup generated by (1,1). Following through with the computation we obtain the answer:

| Z/2Z X Z/4Z | = |Z/2Z| * |Z/4Z| = 2*4 = 8
|H| = 4

8 = [G:H] 4
[G:H] = 2

Thank you very much. I've also iterated all the elements. If I was not mistaken , these two cosets are itself and { (0,1),(1,2),(0,3),(1,0)}.