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What is this abelian group?

  1. May 4, 2007 #1
    1. The problem statement, all variables and given/known data
    G=(Z+Z+Z)/N where Z denote the integers and + is direct sum and
    N = <(7,8,9), (4,5,6), (1,2,3)> or the smallest submodule of Z+Z+Z containing these 3 vectors.

    How would you describe G?



    3. The attempt at a solution
    N = {a(7,8,9)+b(4,5,6)+c(1,2,3)|a,b,c in Z} = {(7a+4b+c, 8a+5c+2c, 9a+6b+3c) | a,b,c in Z}

    For any element in Z+Z+Z such as (1,1,0)
    (1,1,0) + N = {(7a+4b+c+1, 8a+5b+2c+1, 9a+6b+3c)} But each component of the vector in the set must be of the form {(7a+4b+c, 8a+5c+2c, 9a+6b+3c) | a,b,c in Z}. So (1,1,0) + N = (0,1,0)+ {(7a+4b+c, 8a+5b+2c, 9a+6b+3c)| all a,b,c in Z}

    Is the above correct? If it is then feeding some random vectors, i.e (,,)+N seems to always produce {(0,a,b)|a in Z2 and b in Z3}=G where Z2 is Z mod 2. Z3 is Z mod 3.
     
    Last edited: May 4, 2007
  2. jcsd
  3. May 4, 2007 #2

    matt grime

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    I think you mean (Z+Z+Z)/N, and when I first read that I thought N was going to be an integer, and Z/N a cyclic group of order N - sometimes defining things in a certain order leads to confusion.

    It is certainly not true, for the reasons you wrote, that for any element g in Z+Z+Z that g+N<N. If it were the case then N would be equal to Z+Z+Z (I don't think that is the case).

    Pick three generators of Z+Z+Z, quotienting by N induces relations on them, that gives a presentation of G.
     
  4. May 4, 2007 #3
    I have corrected my OP and have reviewed by solution to the problem. Is it now correct?
     
  5. May 4, 2007 #4

    Hurkyl

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    Hrm, that's not what I get. (Of course, I could be wrong) This is a good problem for doing some good old-fashioned linear algebra: your kernel is the the rowspace of

    [tex]
    \left(\begin{array}{ccc}
    7&8&9 \\
    4&5&6 \\
    1&2&3
    \end{array}
    \right)
    [/tex]

    and you can do row operations to simplify it. (Just make sure you do row operations defined and invertible over the integers! i.e. the elementary matrix corresponding to it must be an integer matrix of determinant 1 or -1)
     
  6. May 4, 2007 #5
    I have a feeling that what I posted in my OP is wrong.

    I don't understand presentations.

    The presentation in this case which is G = <a,b,c |7a+8b+9c=0, 4a+5b+6c=0, a+2b+3c=0> where a,b,c are the basis vectors for R^3.

    However what does G look like in this form?

    One way to look at G is G=X/(q:X->X)

    where X is R^3. q is the linear transformation given by the transformation matrix A=[tex]
    \left(\begin{array}{ccc}
    7&4&1 \\
    8&5&2 \\
    9&6&3
    \end{array}
    \right)
    [/tex]

    I don't know what G looks like either in this form.
     
    Last edited: May 4, 2007
  7. May 4, 2007 #6
    I realised that you can calculate an equivalent matrix to A which is diagonal by doing row and column operations. In other words finding two basis for the linear transformation to occur.

    So G is isomorphic to X/(p:X->X) where p is a transformation given by the diagronal matrix D which is equivalent to A.

    D happens to be
    1 0 0
    0 3 0
    0 0 0

    Since D is diagonal and maps X -> X we can 'break' G up to a sum of cyclic groups quotienting out the transformation for each component. So G is isomorphic to Z/(1) + Z/(3) + Z/(0) = Z/3 + Z

    So now we can finally desribe G with some intuition? It is a submodule {(0,b,c) |b,c in Z and b is mod 3} of R^3.
     
  8. May 4, 2007 #7
    However I still have trouble intepreting the presentation, G = <a,b,c |7a+8b+9c=0, 4a+5b+6c=0, a+2b+3c=0> where a,b,c are the basis vectors for R^3. Is it saying that (7,8,9), (4,5,6) and (1,2,3) are the identity elements in G? However, how can you have 3 identity elements?
     
  9. May 5, 2007 #8

    matt grime

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    They are all equal to the identity. There are not 3 different identity elements. You have three different ways to write the identity.
     
  10. May 5, 2007 #9

    matt grime

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    It looks precisely like that. Groups do not have unique, or canonical presentations. It is known to be impossible, in a reasonable sense, to decide if two groups defined as generators and relations are isomorphic. All you can do is put it in the form that is preferred in your class, and for which you'll be taught the method.
     
  11. May 5, 2007 #10

    matt grime

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    You can't write that. You canot write it as a subgroup of ZxZxZ when it isn't one. The group is Zx(Z/3Z).
     
  12. May 5, 2007 #11
    ok. So G = <a,b,c |7a+8b+9c=0, 4a+5b+6c=0, a+2b+3c=0> where a,b,c are the standard basis vectors for Z^3.

    If (8,6,4) is in G then what would it mean written soley wrt to the standard basis vectors for Z^3? Or what would (8,6,4) in G look like in Z^3?
     
    Last edited: May 5, 2007
  13. May 5, 2007 #12

    matt grime

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    What is R^3? Where did that come from?

    You can't write a,b,c are in G if you then say they are not. What are they? If a'=(1,0,0), b'=(0,1,0) and c'=(0,0,1) in ZxZxZ, then a,b,c are their images in the quotient group G, I presume is what you meant to write.
     
  14. May 5, 2007 #13
    It should be Z^3. I don't understand the representation group G very well. WHat would (8,6,4) in G be? If there are 3 different vectors which all act as 0 in G.
     
  15. May 5, 2007 #14

    StatusX

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    The idea is that you change the basis of Z^3 so that the vectors your modding out become simple. For example, if you had the simpler case G=Z^2/<(2,2)>, you would change the basis from {(1,0),(0,1)} to {(1,0),(-1,1)}, ie, send (a,b) to (a,b-a) (which is easily checked to be an isomorphism). Then (2,2) is sent to (2,0), so <(2,2)> is sent to <(2,0)>, and we get that G is isomorphic to Z^2/<(2,0)>.

    It's easy to see directly that Z^2/<(2,0)> is isomorphic to (Z/2Z)xZ, but one way to prove it would be to construct a homomorphism Z^2 -> (Z/2Z)xZ with kernel <(2,0)>, eg, send (a,b) to (a (mod 2),b). Then apply the first isomorphism theorem. In fact, you can then go back and compose this with the initial change of basis isomorphism and see that sending (a,b) to (a (mod 2), b-a) has kernel <(2,2)>, so that Z^2/<(2,2)> is isomorphic to (Z/2Z)xZ.

    Then, for example, the element (2,3) would be sent to (0,1), so this is what (2,3) "would be" in G, although you should keep in mind that G is really a set of equivalence classes, and (Z/2Z)xZ is a different, but isomorphic group.
     
    Last edited: May 5, 2007
  16. May 5, 2007 #15

    matt grime

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    Again, that doesn't make sense unless you state what (8,6,4) means. We started in ZxZxZ, then took a quotient, so the image of (8,6,4) in G is just some element. THERE IS NO CANONICAL WAY TO WRITE WHAT IT IS. That is why bases are bad. Write [(8,6,4)] for its equivalence class - you know about this stuff - you have to pick representatives of the equivalence classes - there is no canonical choice to make though.

    There are infinitely many elements in any group that 'act as the identity'. They are all equal, though.

    Let's take an easy example. ZxZ=(x,y)/<2x+y> Let's write [x] and [y] for the images of (1,0) and (0,1). The group is then just <[x],[y] | 2[x]+[y]=0>. So things in it are words in [x] and [y], such as 4[x]+17[y], but subject to the relation that [2x]+[y]=0, so [4x]+[17y]=15[y].

    You do understand presentations - they just tell you how to cancel things.
     
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