Left Cosets of Cyclic Subgroup Question from a GRE Math book that I'm having trouble understanding:
Find the number of left cosets of the cyclic subgroup generated by (1, 1) of $$Z_{2} \times Z_{4}$$ where Zn denotes the cyclic group of {0, 1, 2, ..., n - 1} under addition modulo n. 
My understanding:
The explanation in the book states that $$Z_{2} \times Z_{4}= \{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)\}$$ which I see. However, it goes on to say that the cyclic subgroup <(1, 1)> is given by $$<(1,1)> = \{(1,1),(0,2),(1,3),(0,0)\}$$
It is at this point that I am not fully understanding the result. I see that the 1 & 3 in (1, 1) and (1, 3) are equivalent under additional modulo 2 and same for the 2 and 0 (0, 2) and (0, 0), but I'm not understanding how the two pairs of elements relate to one another. Is it that in modular addition of elements broken up into components, we can break 2 into 1 + 1 and add those ones to the first and second components separately? 
The explanation then goes on to say that since $$(0,1)+<(1,1)>= \{(1,2),(0,3),(1,4),(0,1)\}$$ there are two left cosets. Am I correct in understanding that this is due to the fact that there are only two elements in Z2 × Z4 that can be added to the cyclic subgroup <(1, 1)> that result in proper subgroups of Z2 × Z4 (those elements being (0, 0) and (0, 1))? 
 A: First, recall that in a direct product such as $Z_2 \times Z_4$, addition is done componentwise: so $(a,b) + (c,d) = (a+c, b+d)$. Let's apply this to find $\langle (1,1) \rangle$.
$$(1,1) + (1,1) = (2,2)$$
Now the first component is taken modulo 2, so it becomes zero. The second component is taken modulo 4, so it is unchanged:
$$(2,2) = (0,2)$$
Now let's add $(1,1)$ again:
$$(0,2) + (1,1) = (1,3)$$
This time the first component is less than $2$ and the second is less than $4$, so we don't have to reduce either one.
One more time:
$$(1,3) + (1,1) = (2,4)$$
The first component $2$ is taken modulo $2$ so it becomes $0$. The second component $4$ is taken modulo $4$ so it also becomes $0$. Thus $(2,4) = (0,0)$.
Thus we have shown that
$$\langle (1,1)\rangle = \{(1,1), (0,2), (1,3), (0,0)\}$$
Note that this subgroup contains four elements, and the entire group $Z_2 \times Z_4$ contains $2 \times 4 = 8$ elements, so we know that there must be two cosets of $\langle (1,1) \rangle$. One is $\langle (1,1) \rangle$ itself. The other is $\langle (1,1) \rangle$ plus any element not in $\langle (1,1) \rangle$. All of the following are equally valid ways to specify the second coset:
$$(0,1) + \langle (1,1) \rangle$$
$$(0,3) + \langle (1,1) \rangle$$
$$(1,0) + \langle (1,1) \rangle$$
$$(1,2) + \langle (1,1) \rangle$$
$$\{(0,1), (0,3), (1,0), (1,2)\}$$
