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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."