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Normal subgroup thm

  1. Mar 18, 2004 #1
    Can anyone come up with an alternative proof of the following?

    If H, a subgroup of G, has index [G:H]=p where p is the smallest prime dividing |G|, the H is normal in G.

    I'm already aware of one proof, given here
    http://www.math.rochester.edu/courses/236H/home/hw8sol.pdf
    (page 3 - question #3)

    but I'm hoping to find maybe a more straightforward proof.
     
  2. jcsd
  3. Mar 18, 2004 #2

    matt grime

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    That's about the most straight forward proof I know, and the only one that is purely in terms of groups. It isn't elegantly pretty, but such is a lot of finite group theory.
     
  4. Mar 18, 2004 #3

    matt grime

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  5. Mar 19, 2004 #4
    Yeah, that's what I meant. I wanted a more elegant proof. I like the result, but the proof is kind of clumsy. I can't read that file on my current computer. I'll see if the library computers can read it.
     
  6. Mar 19, 2004 #5
    I went to the Berkeley webpage and found a "dvi" version of the file and my computer liked that better. I really like the theorem presented in that paper, and the proof is just about as long as the less general smallest prime theorem above, and much less clumsy.
    Thanks for your help.
     
  7. Mar 19, 2004 #6

    matt grime

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    A computer that doesn't like ps but is ok with dvi? What brand of *nix are you using? For the future, try ggv, ghostview, kghostview, or download it and to ps2pdf
     
  8. Mar 19, 2004 #7
    *nix? I don't know!
    Is theorem 2 from the paper useful in proving the normality of groups in any context other than in proving theorem 1? Can you give me an example in which p=[G:H] is not the smallest prime dividing |G| and theorem 2 can be used to prove H normal in G?
     
  9. Mar 19, 2004 #8
    *nix? I don't know!
    Is theorem 2 from the paper useful in proving the normality of groups in any context other than in proving theorem 1? Can you give me an example in which p=[G:H] is not the smallest prime dividing |G| and theorem 2 can be used to prove H normal in G?

    [I don't know why this posted twice! Maybe a bug - maybe I clicked too many times.]
     
  10. Mar 20, 2004 #9

    matt grime

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    He gives an example of this situation in the paper right after the statement of theorem 2. It must be possible as these properties are equivalent to normality, so if any one of them holds then the subgroup is normal.
     
  11. Mar 20, 2004 #10
    Yes, but I can't imagine an instance in which it would be easier to prove one of the other conditions than to just straight-out prove normality (by say the normal subgroup test). His examples showed how the theorem is useful for proving a subgroup is not normal. One of them proved normality, but this seemed to be a rather isolated incident. I was wondering if in fact the theorem could facilitate proving normality of a subgroup in a more general context.
     
  12. Mar 20, 2004 #11

    matt grime

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    So you want to use the theorem to prove something is normal, but simultaneously you don't want to use the theorem because it seems no easier than checking directly if the subgroup is normal?

    I'm not sure I see what you really want. Maybe you'd like to conclude: if H is a subgroup of index p and p satisfies some hypothesis then H is normal in G? There will be no simple criteria of this form just involving the prime dividing the order as the example of S_3 shows - there is a subgroup of index 3 that is not normal, 3 is the largest prime and the second smallest prime dividing the order.

    I would guess that if you give me almost any composite number N I can find a group of order N with a subgroup of any given prime index except the smallest that is not normal by finding some extension of some cyclic group.

    Here's another criterion for normality: a subgroup is normal iff it is the union of conjugacy classes. That's useful for showing A_5 is simple.
     
  13. Mar 21, 2004 #12

    matt grime

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    How about this counter example (i've not checked all the details.

    Let N=pqr

    p<q primes, and r what's left over.

    let C_s denote the cyclic group of order s.

    C_p and C_q act on the set 1,...q by permuting the elements cyclically.

    let h be in C_p, this some times determines an automorphism of C_p by conjugation (considering them as subgroups of S_q) so you may get an extension of that means there is a non-normal extension so that C_pxC_r is a non-normal extension of index q.


    There are other things you can do with working out normalizers of sylow subgroups.

    I don't see a way to glue these all together to get some useful set of rules - just some more isolated examples.

    I guess part of the problem is that, say for N=6, there are two subgroups of order 6, and both have subgroups of index 3. In one case it is normal, in the other it isn't. So the structure of the group will have to come into it.


    Try posting to sci.math, you might get a response from derek holt (mareg@mimosa....) who'll know some answers
     
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