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Group Presentations - do they determine the group

  1. Jan 23, 2012 #1
    I am seeking to gain a good understanding of group presentations

    Currently I have the following general question:

    "Does a group presentation completely determine a particular group?"

    The textbooks I am reading seem to indicate that a group presentation does actually determine/specify the group.

    For example on page 31 of James and Liebeck: Representations and Characters of Groups we find:

    "Let G be the dihedral group [itex] D_{2n} = D_8 = <a,b: a^4 = b^2 = 1, b^{-1}ab = a^{-1}> [/itex]

    Is this a complete specification of the dihedral group - i.e. does this presentation completely determine or specify the dihedral group [itex] D_8 [/itex]?

    Surely it does not - because we additionally need to know that (or do we?)

    a = (1 2 3 4) [ rotation of a sqare clockwise through the origin - see attached]

    and

    b = (2 4) [reflection about the line of symmetry through vertex 1 and the origin - see attached]

    Possibly we also need to know that the elements of the group are

    [itex] D_8 = \{ 1, a, a^2, a^3, b, ba, ba^2, ba^3 \} [/itex]

    but I suspect this can be deduced from the given relations.
     

    Attached Files:

    Last edited: Jan 24, 2012
  2. jcsd
  3. Jan 24, 2012 #2

    morphism

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    Yes, certainly. A presentation <S|R> with generators S and relations R is just shorthand for the quotient of the free group F(S) on S by the normal subgroup generated by R - in particular, it's a group. So the equality G=<S|R> is an equality of groups (or, maybe more precisely, an isomorphism of groups).
     
  4. Jan 24, 2012 #3
    So, just to re-confirm this - a presentation completely specifies a group?

    We do not, in the case of the group [itex] D_8 [/itex] even need to know the nature of the genarators a and b?
     
  5. Jan 24, 2012 #4

    lavinia

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    [/itex]

    The presentation defines the relations between the generators of the group. This determines the group up to isomorphism.

    If the group,G, has n generators, then there is a homomorphism from the free group on n generators onto G that takes each generator in the free group to the corresponding generator in G. The kernel of this homomorphism is the complete set of relations in G.

    If H is isomorphic to G then the group of relations is the same.


    Take your example of the dihedral group of order 8.
    It is the group generated by a rotation of the plane by 90 degrees and a reflection about the y axis.

    Now look at the group of linear transformation of R^4 generated by the two matrices

    0 0 0 1
    1 0 0 0
    0 1 0 0 A
    0 0 1 0

    0 0 1 0
    0 1 0 0
    1 0 0 0 B
    0 0 0 1

    This group is isomorphic to the dihedral group of order 8.

    This is because A[itex]^{4}[/itex] = B[itex]^{2}[/itex] = Id and BAB = A[itex]^{3}[/itex]


    As you can see, even though this is a different group it has the same presentation as the first group.
     
    Last edited: Jan 24, 2012
  6. Jan 24, 2012 #5
    Hi Lavinia

    Thanks for the help

    Much appreciated

    Peter
     
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