I have some questions about groups presentations.

I think I understand what a group presentation is.

Take a set A, form the free group F[A], take a set of relators R in F[A], and from the smallest normal subgroup of F[A] containing all the relators. This is the subgroup <R> of F[A] generated by the conjugates of the relators. (I'm not sure about this notation. This is called the normal closure of R. Right?) Form the factor group F[A]/<R> and you get some group sort of like F[A], except all elements of <R> are collapsed to the identity.

For example, if A = {a} and R = {a^6}, then F[A]/<R> is isomorphic to Z6 since each element of F[A] of the form a^(6k) is collapsed to the identity. So a presentation for Z6 is

(a: a^6) using the relator notation or (a : a^6=1) using the relation notation.

Here is my question:

Given a presentation, how can I determine whether or not all possible elements of the set defined by the presentation are distinct elements of F[A]/<R>?

For example. Given

(1) (a,b: a^5=1, b^2=1, ba=(a^2)b)

(2) (a,b: a^5=1, b^2=1, ba=(a^3)b)

(3) (a,b: a^5=1, b^2=1, ba=(a^4)b),

each of these presentations defines a group which is contained in the set

{(a^0)(b^0), ab^0, (a^0)b, ab, (a^2)(b^0), (a^2)b, (a^3)(b^0), (a^3)b, (a^4)(b^0), (a^4)b}.

It is possible to show in (1) that a=1 by using the relations:

a=bba=b(ba)=b(a^2)b=(ba)ab=(a^2)bab=(a^2)(ba)b=(a^2)(a^2)bb=(a^4)(b^2)=a^4.
So a^3=1. a^3=1 and a^5=1 imply a^2=1. a^2=1 and a^3=1 imply that a=1.

Hence (1) gives just {1,b}; a group isomorphic to Z2. Similarly, (2) also gives a group isomorphic to Z2. But apparently (3) does give a group with exactly 10 elements. How can I show this? I assume I use contradiction, but what kind of contradictions can I look for? Is a=b a contradiction? Is a=1 or b=1 a contradiction since in the free group the identity is the empty word? I guess I just don't know what I can assume about a and b.

By the way, all these examples come straight out of Fraleigh's "Abstract Algebra."

Hurkyl
Staff Emeritus
Gold Member
(a,b: a^5=1, b^2=1, ba=(a^4)b),
...
I guess I just don't know what I can assume about a and b.
You can assume {a,b} generates the group, that a^5=1, that b^2 = 1, and that ba = a^4b. You can assume no further equations hold, unless they can be proven from these three.

Incidentally, you've already enumerated the elements of your group (possibly with repetition). You've described a general calculation of computing <R>, and then taking the factor group formed by modding out by <R>. Since you're dealing with a small, finite set, this should be fairly straightforward to do....

Could you give me an example? For instance, why is it true that ab is not equal to a. Is it because ab=b implies a=1. If so then why is this a contradiction?

Hurkyl
Staff Emeritus