How do I *formally* get informations from a presentation for a group? Just for clarification, here is the definition for a presentation for a group:

Let $G$ be a group and $S$ be a set.
Let $F(S)$ be the free group on $S$ and $R\subset F(S)$ and $\overline{R}$ be the normal closure of $R$.
Then, $(S|R)$ is a presentation for $G$ if and only if $G\cong F(S)/\overline{R}$.
(And for convenience, we write $r=1$ if $r\in R$)

With this definition, let's consider an example, $(x|x^6=1)$ is a presentation for $\mathbb{Z}_6$. BUT Why?
To make that assertion, I have to show the existence of an isomorphism $\phi:F(\{x\})/\overline{\{x^6\}}\rightarrow \mathbb{Z}_6$ and THIS IS NOT TRIVIAL to me.
Even with a really basic group, I have a trouble with visualizing a presentation. I cannot even guess what would presentations would look like fo complicated groups.
How do I formally get informations from a presentation? Please help me with details..
This is how I feel what others do: One concludes $(x|x^6=1)$ is a presentation since $\mathbb{Z}_6$ is cyclic so that generated by a single element and since $6•1=0$. This doesn't really seem legit to me.
Here's an illustration how I view this: By the definition of free group, we know that $\mathbb{Z}$ is a free group with a basis $\{1\}$. Since $|\{1\}|=|\{x\}|$, $\mathbb{Z}\cong F(\{x\})$. Now, we have to take the normal closure of $\{x^6\}$.. Hmm what would the closure look like...?
 A: To get an intuition of what a presentation is saying, maybe it helps to think about the names we give: generators and relations.
You have a set of generators (and their inverses). The idea is that they should generate that group: any element can be written as a product of the generators, a bit like any element of a vector space being a linear combination of the basis vectors.
The problem is, that basis vectors don't interact (that's the point of linear independence), but you can't get away with that in general for groups. So if you just take generators, and don't say anything about them (other than that they cancel with their inverses) you only get free groups. Free groups are groups with as little interaction between elements as possible.
To get more types of groups, we therefore specify relations, telling us how the generators interact with each other. There will in general be infinitely many actual relationships, but we don't want to write them all down. Instead, we write down a (hopefully) meaningful set of relations, that 'generate' all the real ones. And what we write down is which elements are in fact the same.
So, for example, take your presentation $(x|x^6=1)$. That says we take one element ($x$), and we start multiplying it by itself. This gives us $x, x^2,x^3,x^4,x^5$. But when we get to $x^6$, we declare that we've in fact got back to $1$. So we have an element of order $6$, and every element of the group is a power of that element. That is the definition of the cyclic group of order $6$.
To get $\mathbb{Z}_6\times\mathbb{Z}_7$, we take the two things we know: $(x|x^6=1)$ and $(y|y^7=1)$. We know everything we want is given by combining these, so we take the two generators $x$ and $y$. As $x$ and $y$ still interact with themselves as before, the relations $x^6=1$ and $y^7=1$ still hold. But we also need to say something about how $x$ and $y$ interact. By definition of the product, we want $x$ and $y$ to commute, so we add in the relation $xy=yx$. In fact, that is enough, so $\mathbb{Z}_6\times\mathbb{Z}_7$ has a presentation $(x,y|x^6=1,y^7=1,xy=yx)$.
To get a more visual understanding, find out about Cayley graphs.
