Free Groups, Relations and Presentations

Quotient and Kernel

Every group G is the homomorphic image of a free group. Let S be a set of generators for G, and let F be the free group on S. Now F wraps around G, with the words of F mapping to the corresponding words in G. Since S generates G, F maps onto G, and the homomorphism is an epimorphism.

The kernel of this map is a normal subgroup of F. These are the words that produce e in G.

If a set of strings generates the kernel, these strings are called relators, and they necessarily map into e in G. Specifying S and the relators is sufficient to describe G, up to isomorphism.

Conversely, given any free group F on S and a set of relators Y, we can construct a normal subgroup and hence a quotient group G. Let H be the intersection of all normal subgroups containing Y. There is at least one normal subgroup, namely F, and the intersection of normal subgroups is normal, so H is well defined, and F/H = G.

Presentation

In the above example, S and Y form a presentation for the group G. A presentation consists of a set of generators, and a set of relators, words in the free group on S. The group defined by this presentation is the free group mod the normal subgroup implied by the relators.

In fact, G is the largest group that carries the relators to e. Assume S generates another group G′, whose kernel, H′, contains Y. Since H′ contains H, an epimorphism carries F/H onto F/H′. thus G maps onto G′. In some sense G is a universal object in the category of groups generated by S, with Y in the kernel.

Abelian groups and modules can also be described using presentations.

A finite presentation has finitely many generators and relators.

Relators and Relations

In the above I was careful to use the word relator, rather than relation, although these terms can often be used interchangeably. A relater is a word, an element in the free group. A relation sets one word eequal to another. This is of no consequence when dealing with groups, since w₁ = w₂ is the same as w₁/w₂. However, there is a category where the distinction is important.

A monoid is almost a group, but it doesn't have to have inverses. The free monoid on S consists of words formed from the letters of S. There are no inverse letters, and there is no cancellation. Consider the free monoid on A and B, with the relation A = B. There is no relator A/B, because there is no B inverse. In fact the quotient monoid cannot be produced using relators, because the kernel is empty. A and B both map to A, BB maps to AA, and so on. Only e maps to e, so there aren't any relators generating a kernel.

Monoids are not a typical category. When dealing with groups, abelian groups, or modules, one can specify relations or relators, which are then used to build the kernel for the quotient map. As mentioned earlier, relators are usually called relations, with no real distinction between the two.

Normal Closure

In the category of modules, the relators generate a submodule that becomes the kernel for the quotient map. The situation is a little more complicated with groups, because the subgroup generated by the relators may not be normal in F.

Let Y be a set of relators, and expand Y to include wx/w for every relator x in Y and every word w in F. The normal closure of Y has to include wx/w, and when all these conjugates are folded into Y, the subgroup H, spanned by Y, is normal in F. For instance, take v from F and conjugate w₁x₁/w₁*w₂x₂/w₂, which is vw₁x₁/w₁/v * vw₂x₂/w₂/v, which is generated by Y.

Of course it is not practical to write down wx/w for every w in F, but that's the idea.

The quotient group is often easier to understand when relators are turned into relations. If you are given the relation A = B, you can replace B with A wherever it appears in the quotient group. This relation is equivalent to the relator A/B, and every conjugate thereof. Assume uBv is in the quotient group. Multiply on the left by u*(A/B)/u, an element in the kernel, and get uAv.

It is sometimes convenient to multiply a relation on the left or right by a letter or word. The result is another relation that must hold trure. In the above example, uBv = uAv is a valid relation, though it was not part of the original presentation. This trick will be used in the next example.

Example

Consider the presentation with generators A and B, and the following relations.

AB/A = B²
BA/B = A²

Multiply the first by 1/B on the right, and invert the second.

AB/A/B = B
B/A/B = 1/A²

Substitute the second into the first, and 1/A = B, or AB = 1. Plug this into the original relation and 1/A = B², hence B = B², and B = 1. By symmetry A = 1, and the group is trivial.