Diagonally embedded Z4 in direct product of Z8 and Z2

Here, the group $$G$$ is the group direct product of Z8 and Z2, written for convenience using ordered pairs with the first element an integer mod 8 (coming from cyclic group:Z8) and the second element an integer mod 2. The addition is coordinate-wise.

$$G$$ has 16 elements:

$$\! \{ (0,0), (1,0), (2,0), (3,0), (4,0), (5,0), (6,0), (7,0), (0,1), (1,1), (2,1), (3,1), (4,1), (5,1), (6,1), (7,1) \}$$

The subgroup $$H$$ is:

$$\! \{ (0,0), (2,1), (4,0), (6,1) \}$$

$$H$$ is isomorphic to cyclic group:Z4 and the quotient group $$G/H$$ is isomorphic to cyclic group:Z4 as well.

Cosets
$$H$$ has four cosets in $$G$$ (it is a normal subgroup since $$G$$ is an abelian group, so the left and right cosets coincide):

$$\! \{ (0,0), (2,1), (4,0), (6,1) \}, \{ (1,0), (3,1), (5,0), (7,1) \}, (2,0), (4,1), (6,0), (0,1) \}, \{ (3,0), (5,1), (7,0), (1,1) \} $$