Problem: Show that the group defined by the presentation 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 are distinct. But to do this we need to prove that there is no sequence of basic operations on the constraints that will allow us to conclude, for example, that . 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 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 on a set can be defined loosely as the group of words in elements of , where the group operation is concatenation, taking care to add inverses and enforce relations like .

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

(let me know if you know how to make less ugly LaTeX diagrams in WordPress) Here is simply the natural inclusion map, and by commutes we just mean that .

How can we use this? Well, our original group can (and should!) be defined as a quotient , where is the free group on and is the normal closure of . It would help if we could construct a surjective homomorphism from onto some infinite group , (you could call this a realization of the group) and indeed, the universal property allows us to do this.

One such realization is , in . (this is the projective special linear group) These matrices are chosen so that their product has infinite order.

What we would like to say is that there exists a homomorphism from to that maps and 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 by , , and apply the universal property to produce a map that maps to and to .

If we can show that , then will naturally induce a map . But this is easy to verify: , (remember that and generate ) so we have now produced a map such that has infinite order in . It follows that 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 . 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 is given by a presentation and a group has generators consistent with that presentation, is a homomorphic image (hence a quotient) of . 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.

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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.