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A group of order 36 is simple

  1. Dec 7, 2011 #1
    Well it isn't.
    I'm trying to prove it.

    So I assume there is more than one sylow-3 subgroup, each has order 9, we have 4 of them, now their intersection is either e or has order 3.

    if it is e, then we have 32 elements in these subgroups besides e (33rd)

    then assume we have more than one sylow-2 subgroup, we should have 3, but then their intersection has order 1, or 2.

    So now it works for all cases except when the intersection of the sylow 3 subgroups is 3 and that of the sylow 2-subgrps is 1, then I get the magic number 36.

    What should I be looking for in this proof. I was advised there is a way to prove it using this way without the appeal to homomorphisms to Sn.
     
  2. jcsd
  3. Dec 7, 2011 #2

    mathwonk

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    if you check out my math notes for 843-4-5 on my website you will find proofs that all groups of non prime order n different from 60 are not simple for n < 168.


    843-1. #9.

    http://www.math.uga.edu/%7Eroy/843-1.pdf [Broken]
     
    Last edited by a moderator: May 5, 2017
  4. Dec 7, 2011 #3
    Thanks, I am going to check it now.
    btw I corrected my answer to the intersection of sylow p-subgroups. I didn't know that there's a theory on these. The Sylow intersection.

    so the ord(sylow intersection) divides the original group but must NOT be equal or larger to the order of any sylow-p-group (which is the same for all since they are all isomorphic).
     
    Last edited by a moderator: May 5, 2017
  5. Dec 7, 2011 #4
    when it comes to group theory, Hungerford's introductory textbook is truly weak. Not to confuse with his Graduate textbook Algebra though.
     
  6. Dec 8, 2011 #5

    Deveno

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    suppose we have the "bad case" where we have the intersection of the sylow 3-subgroups is non-trivial. let's call one of these sylow 3-subgroups P.

    this means that [itex]\bigcap_{x \in G}xPx^{-1}[/itex] is a subgroup of order 3. show this is a normal subgroup of G, and then G is not simple.

    there is also another way:

    suppose we have 2 sylow-3-subgroups H,K with non-trivial intersection. what are possible orders for N(H∩K)?

    (first rule out 3 and 9. from the first sylow theorem we know there is a subgroup of G of order 9, that H∩K is normal in. this rules out 3. now rule out 9, since...?)
     
    Last edited: Dec 8, 2011
  7. Dec 9, 2011 #6
    I'm inclined to think that our subgroup [itex]\bigcap_{x \in G} \ xPx^{-1}[/itex] is nothing but the original intersection subgroup. The reason: xPx-1 is nothing but another sylow-3-subgrp. Thus the order is 3. Also since for all x in G, if we take [itex]x \bigcap x^{-1}[/itex] then we end up in the same grp hence it's normal.
     
  8. Dec 9, 2011 #7
    How can we have 2 Sylow 3-subgroups and [itex]2 \not \equiv 1 \mod{3}[/itex]
     
    Last edited: Dec 9, 2011
  9. Dec 9, 2011 #8

    Deveno

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    no, i don't mean we have ONLY 2. i mean suppose we have 4. pick any two of them, call them H and K. they have a non-trivial intersection. what is the order of the normalizer of this intersection?

    (hint: any subgroup of G of order 9 is abelian. thus H∩K is normal in both H and K, which means that |N(H∩K)| > 9. since both H and K are in N(H∩K), this means that 9 divides N(H∩K). so we have 9 divides |N(H∩K)| and |N(H∩K)| divides 36. what possibilities does this give?)

    the group [itex]\bigcap_{x \in G} \ xPx^{-1}[/itex] is called the normal hull of P. it is always a normal subgroup of G. see this thread: https://www.physicsforums.com/showthread.php?t=555966
     
  10. Dec 9, 2011 #9
    First let me ask this please, regarding the normalizer, is there a formula for its order?
     
  11. Dec 10, 2011 #10

    Deveno

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    in general, no. but for any subgroup H of G, we have H ≤ NG(H) ≤ G. so normalizers can be used when we need to find "bigger" subgroups of G containing H.
     
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