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More group theory (or module theory)

  1. Jan 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Given an abelian group G with generators x and y, and relations 30x + 105y = 42x + 70y = 0, show it's cyclic and give its order.


    2. Relevant equations



    3. The attempt at a solution
    I'm guessing the proof basically involves cleverly adding 0 to 0 to show that x = ry or y = rx for some r in the integers, but I'm apparently not clever enough to find the trick. I did notice that 30 = 2*3*5, 105 = 3*5*7, 42 = 2*3*7, and 70 = 2*5*7 which is interesting, but I'm not sure how to tie it together, probably due to my weak background in number theory. Any hints?
     
  2. jcsd
  3. Jan 21, 2008 #2
    I think I know what you want to do, but gcd(30, 42) =/= 1 and gcd(105, 70) =/= 1. In fact, none of the integers involved are relatively prime.

    Edit: Disappearing post makes me look silly.
     
  4. Jan 21, 2008 #3
    Okay, I think this might work:

    (42x + 70y) - (30x + 105y) = 12x - 35y = 0
    (30x + 105y) - (12x - 35y) = 18x + 140y = 0
    (18x - 140y) - (12x - 35y) = 6x + 175y = 0
    (12x - 35y) - (6x + 175y) = 6x - 210y = 0

    Then 6x = 210y, and x = 35y. Does that work?

    Edit: Oops. Should be +140y on the third line as pointed out.
     
    Last edited: Jan 21, 2008
  5. Jan 21, 2008 #4
    Yes, except you made an error in your third line of equations (which you seemed not to incorporate beyond that). It does look correct to me.
     
  6. Jan 21, 2008 #5
    Okay, so now I have two questions:

    1. How the heck am I going to find the order of y (and hence the group)?

    2. More generally in this kind of situation would one always proceed in the same way I did?
     
  7. Jan 21, 2008 #6
    You have 12x - 35y = 0 from the top, and you have x = 35y from the conclusion. Assuming everything is correct (and I know what I'm talking about), this implies that 11x = 0. I think that may help (but I'm not sure yet). What do you think?
     
  8. Jan 21, 2008 #7
    Yeah, I noticed that too. Since x is assumed to be a generator, would the order of x give me the order of the group?
     
  9. Jan 21, 2008 #8
    From what I remember in these problems, if x,y are generators, then the group would consist of all possible mx+ny where m,n are integers (and you "mod out" when you can via the relations). Since x has order 11 according to what you did, and x = 35y, x can't generate the entire group, only a subgroup. Since gcd(11,35) = 1 it would seem that y has order 385, and that would be your generator. (I'm hoping someone with more expertise here might chime in, because I'm not entirely sure here, but it does seem to work).
     
  10. Jan 21, 2008 #9
    Sorry, but could you clarify why gcd(11,35) = 1 implies that the order must be 385. I mean, I get why the order divides 385, but I can't come up with a justification for what you said.
     
  11. Jan 21, 2008 #10
    Actually, now that I think about it, x could conceivably be zero. I don't know if the question is answerable without assuming the "generators" are nonzero, assuming that's not implied by calling it a generator.
     
  12. Jan 21, 2008 #11

    NateTG

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    Clearly x=y=0 generates a cyclic group of order 1 and satisfies the listed relationships.

    When you get a generator/relation description of the group it's going to be the 'largest' possible group that satisfies the relations.
     
  13. Jan 21, 2008 #12
    I see. Regardless, if x =/= 0, I can see that eliminates 5, 7, and 35 as possible orders for y, but what about 55 and 77?
     
  14. Jan 21, 2008 #13
    Neither can I at the moment. As I said, I'm not entirely sure here, so it's probably just best for me to let someone with more knowledge answer first. Still, I have a feeling the answer lies in that direction. But intuition is a lousy guide as you probably know.

    Edit: this is an answer to post 9.
     
    Last edited: Jan 21, 2008
  15. Jan 21, 2008 #14

    Hurkyl

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    Just because you're doing group theory doesn't mean you should forget linear algebra! You know that a good way to understand a system of equations is to (reversibly) convert it into a simpler form.

    Because the integers are a Euclidean domain, you can always reversibly convert a system of integer equations into row echelon form using integer row operations. (but you usually cannot put it into reduced row echelon form)
     
    Last edited: Jan 21, 2008
  16. Jan 21, 2008 #15
    So I guess if the thing actually had a unique solution in the integers, that would imply that the two generators are actually unique? Interesting.
     
  17. Jan 22, 2008 #16

    Hurkyl

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    It does have a unique solution in the integers: [0, 0]. :wink: However, the solutions in G are more interesting! (And, essentially by construction, G is the "best" group to consider solutions to the given system of equations)


    But I think you're thinking about it the wrong way... the point is that you presentation of G as [itex]G \cong \mathbb{Z}^2 / H[/itex], where H is the subgroup generated by [30, 105] and [42, 70].

    The most straightforward plan (that I know) is to find simpler generators for H. Since H is the row span of the matrix

    [tex]\left( \begin{array}{cc} 30 & 105 \\ 42 & 70 \end{array} \right)[/tex]

    the natural thing to do is to put it in row echelon form. (Hint: the Euclidean algorithm will help here)
     
  18. Jan 22, 2008 #17
    I think I've got you up till putting the matrix in row echelon form using the Euclidean algorithm. Sorry, I'm a little slow when it comes to this stuff.
     
  19. Jan 22, 2008 #18
    Okay, I think I got it (all my previous work seems wrong, at least in the conclusion). If anyone cares to critique the argument:

    One can use the relations to show that 66x = 385y = 0. Then lcm(66, 385) = 2310 = 2*3*5*7*11 is an annihilator for the group (Z-module) G. So G is a torsion Z-module, and is isomorphic to Z/2Z + Z/3Z + Z/5Z + Z/7Z + Z/11Z, which is isomorphic to Z/2310Z. The last part is based on a theorem that says that a nonzero torsion module can be written as a direct sum of its p-primary components.

    So, does that sound right (particularly the theorem I quoted at the end)?
     
  20. Jan 22, 2008 #19

    Hurkyl

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    That doesn't follow. The group presented by
    Generators x, y
    Relations 66x = 385y = 0​
    is actually Z/66Z + Z/385Z, which is a different group. By reducing to that pair of equations, you've discarded whatever nontrivial relationships exist between x and y.
     
  21. Jan 22, 2008 #20
    Yeah, well, the order made me think it was probably wrong, but I thought I'd see if I was just imagining it. Also, I tried doing various row reductions of the matrix as you suggested, but I just kept getting essentially the same relations I posted before (minus the dubious x = 35y). Oh well, the homework is being turned in in 20 minutes, so I guess it's a lost cause. Thanks anyway. If you do have a solution (or more nudges toward a solution) it would be appreciated though.
     
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