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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

Richard
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    Free monoids don't have anything called $x^{-1}$. – Andrew Dudzik Feb 29 '16 at 15:22
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    Because eleemnts of monoids do not in general have inverses whereas elements of of groups do. – Derek Holt Feb 29 '16 at 15:22
  • I'm not sure that I understand the question, but I thought that part of the point of the definition of a free monoid, over an alphabet containing at least the symbols $x$ and $x^{-1}$, is that it allows you to form expressions such as $xx^{-1}$ and $x^{-1}x$, and thus to form assertions such as $xx^{-1} = 1$, which, if you were to collapse all such expressions to the identity from the outset, would collapse into the tautology $1 = 1$. – Calum Gilhooley Feb 29 '16 at 15:25
  • But the point is that imposing any relation makes it no longer a free monoid. – Andrew Dudzik Feb 29 '16 at 15:32
  • Because "free" means "no nontrivial relations". – Andrew Dudzik Feb 29 '16 at 15:35
  • Does it help (or merely make you as confused as me) to consider an informal analogy with the case of polynomials (over $\mathbb{Z}$, say)? One could consider a `free' structure consisting of expressions such as $1 + x + 2 + x^3 + 0 - x^2 - x$, where $x$ denotes an element of a ring containing $\mathbb{Z}$. But one only gets a (free) ring containing $\mathbb{Z}$ if one takes a quotient of this structure by a congruence relation expressing laws that hold in all rings. This is actually more complex than the group example, but because polynomials are already familiar objects, it might help. – Calum Gilhooley Feb 29 '16 at 16:14
  • Just a sidenote: there are ways to create free groups without making use of the equivalence classes on words that you mention. Have a look here. The answer on that question is very nice, but you need some familiarity with categories to understand. – drhab Mar 01 '16 at 07:53

2 Answers2

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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.

J.-E. Pin
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  • @Richard Further to my previous comment (now deleted), see http://meta.stackexchange.com/questions/5221/how-does-deleting-work-what-can-cause-a-post-to-be-deleted-and-what-does-that, in particular this: "You can’t delete your own question when it has an upvoted answer [...]". – Calum Gilhooley Feb 29 '16 at 18:26
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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.

Andrew Dudzik
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