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Finite Simple Group

  1. Aug 6, 2007 #1
    Question: Suppose that [itex]G[/itex] is a finite simple group of order greater than [itex]5[/itex] that contains no elements of order [itex]2[/itex]. Prove that [itex]G[/itex] contains no subgroup of index [itex]5[/itex].

    Attempt: Suppose G has a subgroup H of index 5, i.e., [G:H]=5. Since 2 doesn't divide |G|, 2 doesn't divide |H|. So the order of H (and the order of G) is odd.

    If |H|=1, then |G|=5 x |H|= 5. This is a contradiction because |G|>5. So |H|> 2.

    Now if [itex]|G|=(5^a) m [/itex] where [itex]5 \not | m[/itex], then for [itex]n_5 = \#(Sylow\: 5\: subgroups)[/itex], [itex]n_5 | m [/itex] implies that [itex]n_5[/itex] is odd. If [itex]n_5 =1 [/itex], this contradicts that G is simple. So [itex]n_5 \geq 3[/itex].

    From Sylow's theorem, we also know that [itex]n_5 \equiv 1 \mod 5[/itex]. So the last digit of [itex]n_5[/itex] must be 1.

    Okay, I think this is the correct approach but I don't see any contradiction. Please help...

    Thank you.
     
  2. jcsd
  3. Aug 6, 2007 #2
    I thought that given a finite group G and any subgroup H of G, we have [G:H]=|G|/|H| according to Lagrange's Theorem. Maybe it's my mistake?

    I just looked up Lagrange's Theorem and I think for any subgroup H, the index of H in G is defined as [G:H]=|G|/|H|.
     
    Last edited: Aug 6, 2007
  4. Aug 6, 2007 #3

    matt grime

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    The index of s subgroup is indeed just the number of cosets - there is no requirement for the subgroup to be normal at all.
     
  5. Aug 6, 2007 #4

    morphism

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    What?

    As for the original question, there is a theorem that states: If G is a finite simple group and H is a proper subgroup of G, then |G| divides [G:H]! (factorial).

    This should help you out.

    Another result that could be used is: If G is a finite group and p is the smallest prime divisor of |G|, then any subgroup of index p in G is normal.

    Both results can be proved using group actions.
     
  6. Aug 6, 2007 #5
    @matt_grime. A mistake I made.

    You got it. The last digit must be one. Now [tex]G=5^am[/tex] since [tex]n_5|G[/tex] it means [tex]n_5=5^bc[/tex] where [tex]c|m[/tex] and [tex]b\leq a[/tex]. But [tex]5^bc[/tex] never ends in 1.
     
  7. Aug 6, 2007 #6

    morphism

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    What if b=0 (which it must be, by Sylow)?
     
  8. Aug 6, 2007 #7
    Then we have,
    [tex]c\equiv 1 (\bmod 5)[/tex] by Sylow's third theorem.
    But [tex]c[/tex] is odd because "G has no element of order 2".


    It seems I am missing something, very tired right now. But it makes sense to me now.

    EDIT: Yes, a mistake. Because c=11, is not a contradiction.
     
  9. Aug 7, 2007 #8

    mathwonk

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    duhhh, now i remember a main point of the course i taught last year.:

    if a group G has a sub group H of index n, then there is a non trivial map G-->S(n) given by the action of G on by translation, on the cosets of H.

    hence if G has a subgroup of index 5, there is a non trivial homomorphism G-->S(5).

    But if G is simple it is injective, then G has order less than 60, i.e. G is isomorphic to A(5), which has an element of order 2, contradiction to hypothesis.
     
  10. Aug 7, 2007 #9

    morphism

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    Yeah, that's basically the proof of the |G| divides [G:H]! thing.

    Once you embed G in S(5), you can finish it off differently by concluding that |G|=15, and then:
    1) S_5 doesn't have a subgroup of order 15. Contradiction.
    2) G =~ C_15, which is not simple (because the only simple abelian groups are the prime cyclic ones), or
    3) G has an element of order 5 by Cauchy, and thus a subgroup of index 3 which must be normal (by the second result I posted). So again G cannot be simple.
     
  11. Aug 10, 2007 #10
    Thanks for all your help! I was traveling for a few days but I'm back into the studying mode. Yes, you guys are right. You're absolutely right! :cool:
     
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