Construction of a free group

Question

I am studying Algebraic Topology, and I need to understand the concept of 'free groups'.
I've read the definition ,a free group is a collection of words formed from a set, called a generating set and it satisfies all of the group axioms.
The problem is that i don't understand how does the concatenation exist since there is no operation defined ? i mean , in order to construct a free group we only need a set how could we say there is an operation without even defining it ? I know it's a silly question, but I feel that I'm missing something
I would be grateful if someone could help. Thank you

Answer 1

Just to simplify the problem a little, let's consider the free monoid on an alphabet $A$ (a monoid is a set with an associative operation with an identity, but not necessarily any inverses). The free group works similarly, but there's a little more complexity hidden there due to the presence of inverses.$^1$

The free monoid on $A$ is the set of finite-length strings we can form taking symbols from $A$, equipped with the concatenation operation. This operation takes two strings, and appends the symbols from the second string (in order) onto the end of the first string, to form a new string. The length of the new string is the sum of the lengths of the old string.

For example, if $A = \{a, b, c\}$, then $bcbaab$ would be one string in the free monoid, and $cca$ would be another string. If $*$ is our concatenation operation, then $$bcbaab * cca = bcbaabcca, \quad cca * bcbaab = ccabcbaab.$$ This operation is associative, and the empty string is the identity, so we do indeed get a monoid. As the above example shows, this operation is not commutative.

While it's true that, for example $cca = c * c * a$, it's important to realise that the meaning of $cca$ is not literally $c * c * a$. It is a string. If you like, it is a function from $\{1, 2, 3\}$ to $A = \{a, b, c\}$, or an ordered $3$-tuple $(c, c, a)$, or however you want to encode strings as sets.

We can't just define strings like $cca$ to be $c * c * a$, because this would be a circular definition of the concatenation operation $*$. We cannot define an operation like $*$ without a set of things it acts upon. If this set of things already uses $*$ in its definition, then this makes the definition circular. We have to have a separate notion of a string before we can begin to talk about how to combine them.

So, with this in mind, yes, there is an operation here, and yes, it's nothing but adding new words to old ones.

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$^1$ For your information, the free group differs from the free monoid in a couple of ways. First, the alphabet $A$ is "doubled", as we must include inverses for each symbol. So, we form strings not just from, say, $\{a, b, c\}$, but from $\{a, b, c, a^{-1}, b^{-1}, c^{-1}\}$. We also need to form an equivalence relation between "equivalent" words. For example, $abcc^{-1}$ should be considered the same as $ab$. For this reason, the elements of the free group are not just strings, but equivalence classes of strings.

Answer 2

The group elements are not words in $a,b,c, \dots$, but instead equivalence classes of words in $a,b,c,\dots,a^{-1}, b^{-1}, c^{-1},\dots$. The equivalence relation is generated by all equivalences $$w a a^{-1} w' \sim w w' \sim w a^{-1}a w'$$ where $w,w$ are words and $a$ can be any of the letters. The group operation is defined by $$ [w][w'] = [ww'].$$ Of course work is required to prove that this is a group and satisfies the relevant universal property.