Write out $( \mathbb{Z}_2, +) \times ( \mathbb{Z}_2, +)$ After just learning the basics of group theory and cyclic groups, I would like to write down an example of a Carthesian product of 2 cyclic groups in an attempt to have a more visual representation. I have just learned the theory of this subject and am not familiar with it yet, so bear with me! 
Is it possible to write out  $( \mathbb{Z}_2, +) \times ( \mathbb{Z}_2, +)$ in a bit of detail, perhaps even with some extra commentations or guidelines ?
 A: You are particularly lucky, as you have chosen a finite group, so we certainly can write it out.
First, what does its elements look like? In general, for two groups A and B, the elements are exactly the pairs (a,b), with $a \in A, b \in B$.
What do these pairs look like in our case? Well, since $$Z_2 = \{0,1\}$$ (Considered as a set, i.e we dont care that the formal definition of a group is a pair whose first coordinate is a set and the second coordinate an operation on the set - I only described the first coordinate, assuming that we agree on the operation anyway).
So the pairs we are looking for are things like $$(0,0),(0,1),(1,0),(1,1)$$ Frankly, these are all the elements in the group $Z_2 \times Z_2$! In other words - we have written out the group. (lets call it $G$ from now)
(If you recall, the Cartesian product of two finite sets of size $n$ and $m$ will be a new set of size $nm$, so we expect to find four elements in our new group).
Now that we know what the elements are - lets do some computing! First, what is the identity element in $G = Z_2 \times Z_2$? In general, this is the pair $(0_a,0_b)$ where $0_a$ is the identity in A, and $0_b$ is the identity in B. In this case, this gives us that the identity in $G$ must be $(0,0)$. 
To give you some idea of how the group operations work in this case - let us compute $(1,0) + (1,1)$. We simply work coordinate-wise, following the definition of products of groups. The result is then $(1 + 1, 0 + 1) = (0,1)$. 
I hope this helps!
