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Questions about group presentations

  1. Dec 28, 2008 #1
    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:

    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."
  2. jcsd
  3. Dec 28, 2008 #2


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    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....
  4. Dec 28, 2008 #3
    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?
  5. Dec 28, 2008 #4


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    Equational relations on a group (or a ring, or even a set!) cannot be contradictory. The relevant question is whether or not a = 1 is in the congruence generated by the relations. (Or equivalently, if a is in the normal closure of the relators)
  6. Dec 28, 2008 #5
    OK. That makes sense. So this gives me a way to show that the set does not contain distinct elements; by using the relations to show that two elements are the same. But how do I show that they are all different. Why is it true that ab is not equal to a?
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