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Show every group of order 77 has elements of order 7 and 11

  1. Apr 30, 2009 #1
    Without Sylow's theorems!!

    This was a problem at the end of a chapter on Lagrange's theorem. I know that every subgroup of order 77 is cyclic. But I don't know how to prove this using only Lagrange. Any suggestions?
     
  2. jcsd
  3. Apr 30, 2009 #2
    Wellll why don't you begin by assuming that there do *not* exist elements of order 7 and 11? In this case the group must be cyclic, generated by, say, a (by Lagrange). What is the order of a^11?

    This argument will prove that there must exist at least one element of order 7 OR 11. A similar argument will prove that there must be elements of both orders.
     
  4. May 3, 2009 #3
    This follows from Cauchy's theorem that if p divides the order of a finite group with finite order, then there is a subgroup of order p. This question is however a specific form of the condition as the order of the group can be anything in cauchys theorem, need not be multplication of two primes as in this case.

    Lets work for this case without cauchy:

    First of all the elements of a group of order 77 are either going to have order 1, 7, 11 and 77 as obvious from lagrange.

    If the group has no elements with order 7 or 11 then all its elements can have order 77 or 1. We know that only identity has order 1. So in this case all elements would have to have order 77. As shaggy says, take any such element a (not e), a^11 has order 7 and a^7 has order 11.

    We used: only divisors of 77 are 7 and 11. We didn't use the fact that they are primes but if you try to prove it generally by cauchy's theorem you will need it.
     
  5. May 3, 2009 #4
    That doesn't complete the proof, for it doesn't handle the case where there is an element of order 7 but not 11 and vice versa.
     
  6. May 3, 2009 #5
    edit: wait right it is not right let me think some more
    edit2: ok as matt says this only applies IF the group is cyclic sorry

    What I wrote was if this group of 77 elements is cyclic, then there exists atleast one element which has order 77. Take that element, a then a^11 and a^7 is certainly elements with order respectively 7 and 11.
     
    Last edited: May 3, 2009
  7. May 3, 2009 #6

    matt grime

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    Sorry, but I think you just said the 77 is prime, and wrote down 7 and 11, the prime factors of 77.
     
  8. May 3, 2009 #7

    matt grime

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    Let G be a group of order 77. If G is cyclic we're done. If not, then either the hypothesis holds or all elements have order 7 or they all have order 11 (excluding the identity).

    HINT: if K and L are subgroups of G and |K|=|L|=p a prime then either KnL={e} or L=K.
     
  9. May 3, 2009 #8
    you are right I am just too sleepy thanks for pointing it out
     
  10. Jul 28, 2009 #9
    I am just looking at this again after a long while, does this work?

    Let G be a group of order 77. If G is cyclic we're done. If not, then either the hypothesis holds or all elements have order 7 or they all have order 11 (excluding the identity). Supposing all non-identity elements have order 7, then using your hint we can write G = {e,x[tex]^{1}_{1}[/tex],x[tex]^{2}_{1}[/tex],...,x[tex]^{6}_{1}[/tex],...x[tex]^{1}_{k}[/tex],x[tex]^{2}_{k}[/tex],x[tex]^{6}_{k}[/tex]} for a total of 6k + 1 elements for some integer k. Similarly, if all non-identity elements had order 11 then G would have 10m + 1 elements for some integer m. Since the equations 6k + 1 = 77 and 10m + 1 = 77 have no solutions in the integers, neither of these cases can occur.
     
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