Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a set and $M$ the free monoid over $X$. Then an automorphism $f$ of $M$ satisfies $f(X)=X$ and so $\text{Aut}(M)$ is canonically isomorphic to $\mathfrak{S}_X$.

My Proof: For every word $w\in M$, let $l(w)$ be the length of $w$. We can show by induction on $l(w)$ that $l(f(w))\geq l(w)$. Since the same is true for $f^{-1}$, we have $l(f(w))=l(w)$ and the special case $l(w)=1$ is all we need to finish the proof.

I have almost no formal knowledge of category theory, but I try to think categorically whenever possible. Since the proposition above can be formulated in categorical language, I assumed there would be an easy proof using only the universal property of $M$. But then I realized that the analogous statement for groups is not true. For example, $\mathbf{Z}$, the free group over a one element set, has two automorphisms.

The reason the above proof breaks down is that there we had a (monoid) homomorphism $l:M\rightarrow\mathbf{N}$ with $l(X)=\{1\}$, and such a mapping does not exist if $M$ is the free group over $X$.


Is there a name for (a theory about) the property of categories that the automorphism groups of free objects coincide with the permutation groups of the underlying sets?

Is there a "deeper" reason that the category of monoids has this property and the category of groups does not?

share|cite|improve this question
I think this really is just something special that happens for monoids. In just about any other algebraic theory, it is possible for an automorphism to map an element of the generating set to an element not in the generating set. In some sense, monoids do not have enough freedom. – Zhen Lin Feb 15 '12 at 20:19
up vote 7 down vote accepted

Automorphisms of free algebraic structures are well-studied in examples, but many problems are open, and of course there is no general answer to the question when all automorphisms are just permutations of the free generators.

For groups, see "Combinatorial group theory" (Lyndon, Schupp). For lie algebras and various other examples, see "Free rings and their relations" (Cohn). The title "Automorphisms of a free associative algebra of rank $2$" (Czerniakiewicz) is self-explanatory, similarly "The automorphisms of the free algebra with two generators" (Makar-Limanov) and "On the automorphism group of $k[x,y]$" (Nagata). The group of automorphisms of $k[x,y,z]$ is in current research; see for example "Polynomial automorphisms and invariants" (van den Essen, Peretz) and "The tame and the wild automorphisms of polynomial rings in three variables" (Shestakov, Umirbaev).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.