TITLE: I don't understand what a "free group" is!
QUESTION [33 upvotes]: My lecture note glosses over it really, introduces it and says "well it intuitively makes sense" but I say, nope it doesn't.

Free groups on generators $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ is a group whose elements are words in the symbols $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ subject to the group axioms. The group operation is concatenation.

What do I not understand? Well, to star with, where's the identity? The operation, say I denote it $*$, is $x_1 * x_2=x_1x_2$ yes? How is the identity defined? I mean, $e*x_1=ex_1$ because it's "concatenation"  so I cannot conveniently say $e*x_1=x_1$ and ignore the fact I need to "concatenate" it. These are apparently words, symbols not numbers. The inverse doesn't make sense too, $x_1*x_1^{-1}=x_1x_1^{-1}$ and period. Not $x_1*x_1^{-1}=e$. I mean, I don't even know what $e$ is supposed to be in this supposedly group object so I am left puzzled.
I don't see any mathematics here, concatenation, in other words, is just "lining up the symbols in order." It's not like $1 \times 2 \times 10=20$ but $1 \times 2 \times 20=1220$. 
And another problem. Doesn't the free group have order infinity? It can't be finite can it? Because, say I start with $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ but it must be closed under concatenation. Well, $x_1*x_2=x_1x_2$ already causes an issue because clearly we just created a new element. A new word $x_1x_2$. Continuing this way, we keep adding the newly created words and reach infinity.
And before someone directs me to it, no, wikipedia's page on free groups didn't help me understand this either. 
This bizarre notion is confusing and incomprehensible than ever. Does anyone know the answers to my questions?

REPLY [2 votes]: Most important is the property of a free group: In a free group there are no relationships between elements other than those that you can prove by the axioms of a group. 
For example, $aa^{-1} = 1$ follows from the group axioms and is true in every group, and therefore in a free group as well. 
ab = 1 is possible and will be true in some groups, but it doesn't follow from the group axioms, and therefore isn't true in a free group unless a and b were formed from the generators in such a way that in the product ab all the multiplications cancel each other out (that is ab contains a generator followed by its inverse or vice versa, you can remove it, then you have again a generator followed by its inverse and so on until you are left with 1). Nothing like $Z_{12}$ where 5 + 7 = 0 (or (1+1+1+1+1)+(1+1+1+1+1+1+1) = 0) can happen in a free group.