Definitions of free monoids and free groups

Question

Recall that a word over the alphabet $A\cup A^{-1}$ is a sequence of elements of $A\cup A^{-1}$ where $A^{-1}=\{x^{-1}:x\in A\}$.

The set of all the words over the alphabet $A\cup A^{-1}$ equipped with the operation of concatenation is called the free monoid over the set $A\cup A^{-1}$.

I saw in many books that in order to create a free group we have to introduce the equivalence relation that permits to "delete" the symbols like $xx^{-1}$ and $xx^{-1}$, namely that relation that defines reduced words. The monoid equipped with this relation and the operation of concatenation becomes a group and it is called free group.

Now, why do we need the relation? I mean, if we really need it, why in general we don't use it from the beginning when we speak of free monoids, namely, why a free monoid is defined without this relation?

thank you very much

Answer 1

In addition to the comments of Slade and Derek Holt, which are already a very good answer to your question, let me mention a property required for a free monoid, which holds also on an alphabet of the form $X = A \cup A^{-1}$: every map from $X$ to a monoid $M$ should extend uniquely to a morphism of monoids from the free monoid on $X$ to $M$.

Now take $M = \{0, 1\}$, with the usual product on integers and let $f: \{x, x^{-1}\} \to M$ be the map defined by $f(x) = 0$ and $f(x^{-1}) = 1$. If you extend it to $X^*$ you get $f(1) = 1$ (because $f$ has to a monoid morphism and then should send the identity of $X^*$ on the identity of $M$) and $f(xx^{-1}) = f(x)f(x^{-1}) = 0 \cdot 1 = 0$. Thus if you require $xx^{-1} = 1$, you get $0 = f(xx^{-1}) = f(1) = 1$, a contradiction.

Answer 2

An example may help.

Let $S = \{a\}$, $T=\{b\}$. We can construct the following objects:

• The free monoid on $S$. This is the set of words $\{1,a,a^2,\ldots\}$, and it's isomorphic to the natural numbers $\mathbb{N}$.

• The free monoid on $S\cup T$. This is the set of words in both $a$ and $b$, and we can denote it by $\mathbb{N}\ast\mathbb{N}$. Note that it is non-abelian, as is any free monoid on more than one generator.

• The free monoid on $S\cup T$, modulo the relations $ab=1$ and $ba=1$. This is isomorphic to $\{\ldots , a^{-2}, a^{-1}, 1, a, a^2, \ldots\}$, the free group on $S$, and both are isomorphic to the integers $\mathbb{Z}$.

Note that the last example is not a free monoid: since it is abelian, it would have to be isomorphic to either the free monoid on zero generators (the trivial monoid/group), or the free monoid on one generator. But $\mathbb{Z}$ and $\mathbb{N}$ are not isomoprhic.