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I am reading Berstel's and Perrin's THEORY OF CODES.

Now I have trouble understanding the concept of a (non)free monoid.

Given any set $A$ equipping this set with concatention and the empty word I can make a monoid out of it called the free monoid, denoted by $A^*$.

This definition is pretty straightforward, then given any monoid $M$, $M$ is free is there is a isomorphism between $M$ and a free monoid.

What I don't understand is that since any monoid has a (unique and minimal) set of generators $X$, i.e. $X^*=M$, why is it not always the case that $M$ is free.

It can be proved that a monoid is free if it's minimal set of generators is a code.

Here are two examples of monoids over $A = \{a,b\}$.

  1. $M = \{w \in A^* : |w|_a \equiv 0 \text{ mod } 2 \}$
  2. $N = X^*$, where $X = \{ a, ba, ab\}$

The minimal set of generators of the first example is a code, thus $M$ is free and the minimal set of generators of $N$ is not a code, thus $N$ is not free, but I don't see how $N$ is not free, if $X$ is a set and we take every possible string made out of elements of x and on the other hand why is $M$ free could someone give a isomorphism to a free monoid.

Thanks.

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    I'm not sure I understand your definition of $N$. It seems that it is exactly free monoid of $X$ by your definition? If not, what do you mean by $X^*$? – 2017-01-05
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    @David well, that's what I don't understand either. – 2017-01-05

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A monoid $M$ is freely generated by $X$ if every element of $M$ can be uniquely written as a product of generators from $X$. In your second example $\{a,ba,ab\}$ does not freely generate $N$. For example, the word $abaaba$ can be written in two ways: $ab\cdot a\cdot a\cdot ba=a\cdot ba\cdot ab\cdot a$.

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    So in my 2nd example $X^*$ is not free, right? However, according to the definition of a free monoid, where does it fail to become free? – 2017-01-05
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    You probably is confused with the notation. If $X$ is an abstract set, then $X^*$ denotes the free monoid generated by $X$. Here the elements of $X^*$ are all words over $X$. Further you consider the set $A=\{a,b\}$ and take a subset $X=\{a,ba,ab\}$ in the free monoid $A^{*}$. Now $X^*$ denote not the free monoid generated by $X$, but the submonoid of $A^{*}$ generated by $X$. And this submonoid is not free. – 2017-01-05