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Normal Groups

  1. Dec 9, 2011 #1
    If N is normal in G, is its normal quotient (factor) group G/N normal as well?

    And does it imply that any subgroup of G/N which has a form H/N is normal as well?

    Is a subgroup of a normal subgrp normal as well. It appears that not always.

    Thanks
     
  2. jcsd
  3. Dec 9, 2011 #2

    micromass

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    G/N normal in what?? In order to be normal you need two groups: N normal in G. So G/N should be normal in what.

    Let [itex]N\subseteq H[/itex] with N normal. We have that H is normal in G if and only if H/N is normal in G/N. This is an easy exercise.

    No. A normal subgroup of a normal subgroup is called subnormal. Such a group is not always normal. Can you find a counterexample?? (the dihedral group of order 8 should prove interesting)
     
  4. Dec 9, 2011 #3
    G/N is normal in G. It just seems hard to prove.
    if N is normal in G. for all g in G, gng-1 is in N.
    now if H is in G/N then H = aN for some a in g and not in N. now g(an)g-1 is in H only if it is abelian. what then?
     
  5. Dec 9, 2011 #4

    micromass

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    G/N normal in G?? That makes no sense at all. G/N isn't even a subset of G...
     
  6. Dec 9, 2011 #5
    isn't G/N the union of left cosets of N?
     
  7. Dec 9, 2011 #6
    I'm probably understanding this incorrectly but wolfram says:

    For a group G and a normal subgroup N of G , the quotient group of N in G, written G/N and read " G modulo N ", is the set of cosets of N in G.

    http://mathworld.wolfram.com/QuotientGroup.html
     
  8. Dec 9, 2011 #7
    Again: The elements of the quotient group G/N are subsets of G

    http://people.brandeis.edu/~igusa/Math30/quotient.pdf [Broken]
     
    Last edited by a moderator: May 5, 2017
  9. Dec 9, 2011 #8

    micromass

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    No, it's the set of left cosets of N. Not the union.

    Elements of G/N are subsets of G (not elements of G). So G/N can't be a subset of G.
     
  10. Dec 9, 2011 #9

    micromass

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    Yes. That doesn't mean that G/N is a subset of G.
     
    Last edited by a moderator: May 5, 2017
  11. Dec 9, 2011 #10
    So we know G/N is a group, but then it isn't a subgroup of anything.


    How do you imagine G/N like in space (as a group)?
     
  12. Dec 9, 2011 #11
    so I guess every time I think of quotient groups I need to do so in terms of mappings
     
  13. Dec 9, 2011 #12
    in G/N



    consider:

    f: G ---> G/N ​

    g---> gN​


    f is onto homomorphism.

    now image of Hunder this map is H/N

    I don't know if I can use the 3rd isomorphism thm b/c N is not given to us to be normal. But at least, H/N the image of H is normal because f(aha-1) is in the image(H) and since homomorphism then f(a)*f(h)*f(a)-1 in H/N.
     
    Last edited: Dec 9, 2011
  14. Dec 9, 2011 #13
    please let me know if this is correct.

    Thanks a lot, without you I wouldn't have figured this out.
     
  15. Dec 9, 2011 #14
    let a, b in D4 s.t. |b|=2 and |a|= 4

    then K= {e, a2, b, a2b} is a normal subgroup in D4

    but H = {e, b} is a subgroup in K however not normal.
     
  16. Dec 10, 2011 #15

    Deveno

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    more importantly, H is normal in K (since K is abelian, every subgroup of it is normal in it), and not normal in G.

    normality isn't "transitive".

    there are two (equivalent) ways of thinking of a factor group (quotient group):

    1) as a homomorphic image of the original group.
    2) as a group induced by partitioning by a normal subgroup.

    way number 2 is what involves cosets, which are the equivalence classes under the partition. we require normality of the subgroup, so that we can multiply cosets:

    NaNb = Nab for all a,b in G iff xN = Nx for all x in G.

    if we try to form G/H with H some non-normal subgroup, the sets Hab and HaHb aren't the same for at least one pair (a,b) in GxG. the set of cosets can still be formed, we just don't get a group out of it. try this with D3 or D4 and some non-normal subgroup.

    the equivalence of 1 and 2 is what the first isomorphism theorem is all about.
     
  17. Dec 12, 2011 #16
    thanks all ...
     
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