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Group question

  1. Jun 14, 2005 #1

    quasar987

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    Show that if G is a finite group of even order, then G has an odd number
    of elements of order 2.

    I'd appreciate a tip or two. I really don't see how the order of the elements of a group is linked to the order of that group.
     
  2. jcsd
  3. Jun 14, 2005 #2

    Hurkyl

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    There is a very important theorem linking the order of the elements of a finite group to the order of the group, so I imagine you haven't gotten that far in your class yet!


    I like to experiment when I don't know what to do. Try constructing a group with an even number of elements and an even number of elements of order 2.
     
  4. Jun 14, 2005 #3

    quasar987

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    Well the multiplicative group made up of {e,a,b,c,d,f} where e is the identity, so o(e) = 1. We chose the rest so that o(a) = o(b) = 2 and o(c), o(d), o(f) > 2. I can't think of anything that makes this impossible.

    btw, I'm using this book as a teacher: http://www.math.miami.edu/~ec/book/ and the question is on page 31.
     
  5. Jun 15, 2005 #4

    matt grime

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    I can - the group has order 6. There are exactly two groups of order 6 and neither has exactly two elements of order 2. The group structure prevents this from happening. I
    You must specify what ab is, as well as cd etc and in doing so you'll see that you must create another element of order 2, in one case, or that there can only be one element of order 2 in the other.

    Thnik like this: there is the identity, the elements of order 2 and we can pair up all the other elements like {x,x^{-1}} so the group contains 1+2n+m elements where n is the number of pairs of the form {x,x^{-1]} and m is the number of elements of order 2.

    Now, if a group has an element of order 2 hopefully you know that the group must have an even number of elements (lagrange's theorem). What can you do now?
     
  6. Jun 15, 2005 #5

    quasar987

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    Hi matt grime,

    It seems to me that the equation o(G) = 1+2n+m alone gives the conclusion. If o(G) is even, then necessarily, m is odd, qed. no?
     
  7. Jun 15, 2005 #6

    quasar987

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    I'm stuck again already! :mad:

    I don't understand the proof of the last theorem of pp.31. He jumps directly from a trivial statement to the conclusion of the proof: Let m be the smallest integer with 0 < m < n and [itex]a^m \in[/itex] H. Then m|n and a^m generates H.

    How does this implies that m|n ?
     
  8. Jun 15, 2005 #7
    (Using the same variables as in the book). Suppose a^t is in H. We can write t = qm + r with q, r integers and 0 <= r < m (the division algorithm). Note that a^(mq) = (a^m)^q is in H, and since H was a subgroup, a^(-mq) must also be in H. Thus a^t * a^(-mq) = a^(qm + r) * a^(-mq) = a^(qm + r - mq) = a^r is in H. But r was smaller than m, so r can't be strictly positive (it would break the "minimality" of m). Thus r = 0 and t = qm, i.e. m divides t.
     
  9. Jun 15, 2005 #8

    quasar987

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    And since a^n = e and e is in H, then m|n.

    Strange that evidence for the fact that m|n comes logically AFTER we have proved that a^m generates H, while the book states it first. Anyway, thanks again Muzza.
     
  10. Jun 16, 2005 #9

    matt grime

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    yes, but do you know that o(G) is even? it is but you should state why it must be even.
     
  11. Jun 16, 2005 #10
    Are we reading the same problem? Let me quote the first post:

     
  12. Jun 16, 2005 #11

    matt grime

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    sorry forgot that part of it
     
  13. Jun 17, 2005 #12

    quasar987

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    This question sounds strange: "Suppose G is the additive group [itex]\mathbb{Z}[/itex] and H = 3[itex]\mathbb{Z}[/itex]: Find the cosets of H."

    Doesn't the cosets of H depend on the sub-group one choses to "generate" them? For exemple, if I chose the trivial subgroup H, then the cosets are just H itself. If I chose [itex]H_6 = \{6i |i\in \mathbb{Z}\}[/itex], then the cosets are [itex]H_6[/itex] and [itex]3+H_6[/itex], etc.
     
  14. Jun 18, 2005 #13
    But you were /given/ H, there's nothing to choose there...
     
    Last edited: Jun 18, 2005
  15. Jun 18, 2005 #14

    quasar987

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    Oh, so by "cosets of H" they mean "the cosets making up a partition of G which are generated by H". This is a very bad term.
     
  16. Jun 18, 2005 #15

    matt grime

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    no, it is a good term. 3Z is exactly what you would call H_3. i think you don't understand the terminology since you could do it for 6Z which you called H_6, but think it is bad for 3Z.
     
  17. Jun 18, 2005 #16

    quasar987

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    Could you explain the terminology then please?
     
  18. Jun 19, 2005 #17

    matt grime

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    the cosets of 3Z in Z awhich we'll denote [0], [1], and [2] ie there are three cosets. [0]=3Z, [1]={3n+1, n in Z) and [2] is the set {3n+2, n in Z}

    i fail to see where you're struggling, since it is just a definition - what do you find difficult about the terminology?
     
  19. Jun 19, 2005 #18

    quasar987

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    Nothing, I tought I had understood it, but then you said I didn't understood the terminology:

    But what you said is what I tought it was after Muzza gave his "hint". But thanks for further clarifying that for me matt. I still think it's a bad term though :wink: .
     
    Last edited: Jun 19, 2005
  20. Jun 20, 2005 #19

    matt grime

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    Seeing as you said that you could do the question for "H_6" but not for "3Z" i stand by the fact you don't udnerstand the terminology (ie what 3Z is not what a coset is) since 3Z is H_3 in your terminology. this wasn't about cosets. it was about H.
     
  21. Jun 20, 2005 #20

    quasar987

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    k

    --------------
     
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