Cartesian Product of Sets and the Direct Product of Groups I'm having a bit of confusion. I've tried to search youtube and whatnot but I could not find any explanations. My book says the following:
The Cartesian Product is denoted by: $$S_1\times S_2\times\cdots\times S_n.$$
Also denoted as: $$\prod_{i=1}^nS_i.$$
Now, I know what the cartesian product is from finite math. It is simple to understand that. You take a set $A$ with elements $a$, and a set $B$ with elements $b$. Then the cartesian product of $A\times B=(a,b)$ such that $a\in A$ and $b\in B$. But, now I am looking at the cartesian product in the perspective of modern/abstract algebra. I am confused and I do not particularly understand what this really means. 
For example, how would one go by finding the elements of $\mathbb{Z}_2 \times \mathbb{Z}_4$ as well as the order and determining if this group is cyclic? If we are considering this under addition, which I assume we are, would we simply proceed as follows (for the example $\mathbb{Z}_2 \times \mathbb{Z}_3$ instead):
$$1(1,1)=(1,1)$$
$$2(1,1)=(1,1)+(1,1)=(0,2)$$
$$3(1,1)=(1,1)+(1,1)+(1,1)=(1,0)$$ And we would proceed so on, working modulo 2 and 3, respectively. This is a worked example in my book. The book says that we have six elements, because we have $2\times 3=6$. They claim it is cyclic. Now, I can see if that if they claim it is cyclic, then they only need one generator to generate the whole group and can thus test $(1,1)$. Now, how did they know to pick this specific generator? Would I follow this same process for the question asked above?
I'm also confused of what the direct product of a group G is, as well as the direct sum of a group G.
Edit: I think I have figured out how to find the order as well as the elements. So, if we take my example, we would have order=$2\times 4=8$. Moreover, if we want to find the elements, we take different combinations of $(a,b)$ such that $a\in\mathbb{Z}_2=\{0,1\}$ and $b\in \mathbb{Z}_4=\{0,1,2,3\}$:
$$\implies \{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)\}$$ Now, this is not cyclic because 2 and 4 are not coprime (relatively prime).
 A: One of your questions has been answered here:
Can someone explain the precise difference between of direct sum and direct product of groups?
Multiplication of $G \times H$ is defined as follows.  Given any $(a, b) \in G$ and $(c, d) \in H$, then:
$$(a, b)(c, d) = (a \cdot b, c * d)$$
where $\cdot$ is the operation in $G$ and $*$ is the operation in H.
The order of a group $\mathbb{Z}_m \times \mathbb{Z}_n$ will be $m \cdot n$.  This group is cyclic $\iff$ $gcd(m, n) = 1$.  In general, $|G \times H| = |G||H|$.
If it is cyclic, then $(1, 1)$ is necessarily the generator since the order of any element $(a, b) \in \mathbb{Z}_m \times \mathbb{Z}_n$ will be $lcm(|a|, |b|)$.  This rule holds in general for any $G \times H$.  Of course, remember that the order of a generator must equal the order of the group.
A: When you take the Cartesian Product of two groups $G$ and $H$ you treat their operations separately. So for all $g_1,g_2\in G$ and $h_1,h_2\in H$, the elements $(g_1,h_1)$ and $(g_2,h_2)$ are "composed" to get $$(g_1,h_1)\circ(g_2,h_2)=(g_1\cdot g_2,h_1*h_2)$$ where $\cdot$ is the operation for $G$ and $*$ is the operation for $H$. In general there are $|G|\cdot|H|$ elements in $G\times H$ because there are $|G|$ possible choices for the first entry and $|H|$ possible choices for the second entry in $(g_i,h_j)$.
