Universal Properties I: Free Groups

Problem: Show that the group defined by the presentation \langle a, b | a^2=b^3=e\rangle is infinite.

If you are familiar with the definition of a group presentation, this problem should seem intuitively obvious to you.  But it is remarkably hard to come up with a good proof.

A first idea might be: show that the elements ab, abab, ababab, \ldots are distinct.  But to do this we need to prove that there is no sequence of basic operations on the constraints a^2=b^3=e that will allow us to conclude, for example, that abab = abababab.  It is certainly not immediately clear how to prove this, and a rigorous proof will probably involve several painful lemmas about words from the set \{a, b\} and when they can be equal to the identity.

There is another way  to proceed, which is more abstract and seems to be more standard in algebra, and that is to appeal to the universal property of free groups.  A free group F on a set S can be defined loosely as the group of words in elements of S, where the group operation is concatenation, taking care to add inverses and enforce relations like aa^{-1}=e.

But we can also refer to free groups by the following universal property: If H is any group and \phi: S \to H is any set map, then there exists a unique group homomorphism \Phi such that the following diagram commutes:

\begin{array}{rclc} F \\ \iota \uparrow & \searrow \Phi \\ S & \rightarrow & H \\ & \phi \end{array}

(let me know if you know how to make less ugly LaTeX diagrams in WordPress) Here \iota:S \to F is simply the natural inclusion map, and by commutes we just mean that \Phi \circ \iota = \phi.

How can we use this? Well, our original group G=\langle a, b | a^2=b^3=e\rangle can (and should!) be defined as a quotient F / N, where F is the free group on S=\{a,b\} and N is the normal closure of \langle a^2, b^3\rangle. It would help if we could construct a surjective homomorphism from F/N onto some infinite group H, (you could call this a realization of the group) and indeed, the universal property allows us to do this.

One such realization is a \to \alpha = \overline{\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}}, b \to \beta = \overline{\begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}}, in PSL_2(\mathbb{C}). (this is the projective special linear group) These matrices are chosen so that their product \alpha \beta = \overline{\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}} has infinite order.

What we would like to say is that there exists a homomorphism from G to PSL_2(\mathbb{C}) that maps a and b as above, but we do not know a priori that simply declaring this map will produce a well-defined homomorphism. Here is where the universal property comes in. We can clearly define the map \phi : S \to PSL_2(\mathbb{C}) by \phi(a) = \alpha, \phi(b) = \beta, and apply the universal property to produce a map \Phi : F \to H that maps a to \alpha and b to \beta.

If we can show that N \subset ker\ \Phi, then \Phi will naturally induce a map \overline{\Phi} : F/N \to H. But this is easy to verify: \alpha^2=\beta^3=e, (remember that a^2 and b^3 generate N) so we have now produced a map \overline{\Phi}: G \to H such that \overline{\Phi}(ab)=\alpha\beta has infinite order in H. It follows that ab has infinite order and we are done.

It is important to realize that we can do exactly this sort of realization whenever we can identify a group with generators that satisfy relations consistent with the relations in N. Furthermore, this approach is generalizable to many other situations where universal properties are involved, (tensor products, for example) and helps us avoid the (sometimes messy) definitions of the objects involved.

On an intuitive level, all that has been said here is: whenever a group G is given by a presentation and a group H has generators consistent with that presentation, H is a homomorphic image (hence a quotient) of G. The use of this fact is a key proof technique for problems like the one above, in which we essentially need to prove that a certain structure is not too degenerate.

Possibly related posts: (automatically generated)

This entry was posted on Friday, August 22nd, 2008 at 1:49 am and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

4 Responses to “Universal Properties I: Free Groups”

  1. Keegan Says:
    August 25, 2008 at 3:21 pm | Reply

    Hey,
    If you have xymatrix you can make some pretty good looking diagrams. In this example it would be

    \xymatrix{
    F \ar@{–>}[rd]^\Phi & \\
    S \ar[u]^\iota \ar[r]^\phi
    & H
    }

    xymatrix can be included by \usepackage[all,cmtip]{xy}

  2. lydianrain Says:
    August 25, 2008 at 6:09 pm | Reply

    Keegan, \usepackage doesn’t seem to be handled by WordPress, or else I don’t know how to use it. Could you be more specific?

  3. nbloomf Says:
    August 30, 2008 at 7:58 pm | Reply

    The \usepackage command has to come before the \begin{document} command. I don’t know anything about LaTeX in WordPress, in particular whether you are allowed to include arbitrary packages (from CTAN).

  4. lydianrain Says:
    August 30, 2008 at 8:15 pm | Reply

    My difficulties are specifically WordPress related, thanks.

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