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Prove M is subgroup

  1. May 6, 2010 #1
    Given a group G of order 22 and [tex]M = \{x \in G | x^{11}=e\}[/tex]. Prove that M is a normal subgroup of G.

    I have troubles proving M is subgroup of G. If M was a subgroup, then I can show it is normal, but how to prove it's a subgroup?
    I know I have to show it's closed under multiplication and opposite element, but cannot do this. May I receive some help?
  2. jcsd
  3. May 6, 2010 #2
    1. not precalculus.
    2. Closure. Assume x, y in M. You are trying to prove that xy is in M. All you have to do is write what the two preceding statements MEAN, and then there is one intermediate step. Not too tricky.

    3. Inverse. You know what the inverse of x (in M) is, right? Since this is an element of M, what else can you say about it (in G)?
  4. May 6, 2010 #3

    2. I have to prove that if x, y are from M, then (xy)11=e. If the group was abelian, it's easy - then (xy)11=x11.y11, but now (xy)11=xyxyxy...xy and don't know how to prove it's equal to e.

    3. I'm not sure what's the term in English, but what I understand by the word "inverse" is such an element y, that if x is in G, then xy=yx=e.
    In this problem, if x is in M, the order of x divides 11, so ord(x) = 1 or ord(x)=11.
    x.x10=x11=e, so the inverse of x must be x10.
    We have to show it's in M.
    (x10)11 = (x11)10=e10=e
    Is this right?
  5. May 6, 2010 #4


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    You have to use what you know about the order of G. The prime factorization of 22 is (2)(11). Use this to show that M has a subgroup of order 11. Can it have more than one?
    Last edited: May 6, 2010
  6. May 6, 2010 #5
    But I cannot prove this, because I can't prove M is a group.
  7. May 6, 2010 #6


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    I think that was a typo. Prove that G has a single subgroup of order 11. This should allow you to identify precisely what the elements of order 11 are (and explain why they commute)
  8. May 6, 2010 #7


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    Yes, sorry for the confusion. I meant to say "show that G has a subgroup of order 11."
  9. May 7, 2010 #8
    I proved that G has an element of order 11, but subgroup... :confused:
  10. May 7, 2010 #9
    Think of e, x, x^2, x^3, ..........., x^10

    try to prove this is a subgroup
  11. May 7, 2010 #10
    Here x is of order 11, right?
  12. May 7, 2010 #11
    certainly. It's now easy to find out what (x^m)(x^n), (x^n)^-1 are and that to
    prove they also are in the subgroup.
  13. May 8, 2010 #12
    Thank you :)
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