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Normal Subgroup of G

  1. Jul 31, 2014 #1
    1. The problem statement, all variables and given/known data
    N is a normal subgroup of G if aNa^-1 is a subset of N for all elements a contained in G. Prove that in that case aNa^-1 = N.


    2. Relevant equations



    3. The attempt at a solution
    Given: N is a normal subgroup of G if aNa^-1 is a subset of N for all elements a contained in G. Assume, aNa^-1 = {ana^-1|element n in N}.
    By definition of a normal subgroup, aN is a subgroup of Na for all elements a in G. Then aNa^-1 is a subgroup of Naa^-1 and is = N. And a^-1Na is a subgroup of aNa^-1 and is = N for all elements a in G, when Na= a(a^-1N)a is a subgroup of a(Na^-1)a = aN.

    Is this approach correct?
     
  2. jcsd
  3. Jul 31, 2014 #2

    jbunniii

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    No, ##aN## and ##Na## are not groups at all unless ##a \in N##. If ##N \unlhd G## then it is true that ##aN = Na## (i.e. there is no distinction between left and right cosets). But this is not the usual definition of a normal subgroup.

    Indeed, the definition given in your problem statement is that ##N \unlhd G## if and only if ##aNa^{-1} \subseteq N## for all ##a \in G##. The goal is to prove that this implies that ##aNa^{-1} = N## for all ##a \in G##. You can do this without mentioning cosets at all.

    Hint: ##aNa^{-1} \subseteq N## if and only if ##N \subseteq a^{-1}Na##.
     
  4. Jul 31, 2014 #3
    So if, aNa^-1 is a subset of N if and only if N is a subset of a^-1Na, then element a is contained in N. Then, aNa^-1= N for all elements a in G.
     
  5. Jul 31, 2014 #4

    jbunniii

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    No, you cannot conclude that ##a \in N##. Indeed, ##a## is an arbitrary element of ##G##.
     
  6. Jul 31, 2014 #5
    So if my conclusion is that elements a are present in G, and aNa^-1 = N for all elements a in G, this proves that N is a normal subgroup of G.

    (I realize how I cannot say that element a is contained in N now).
     
  7. Jul 31, 2014 #6

    jbunniii

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    The definition in your problem statement is that ##N \unlhd G## provided that ##aNa^{-1} \subseteq N## for all ##a\in G##.

    The goal is to conclude the apparently stronger statement that if ##N \unlhd G##, then in fact we must have equality: ##aNa^{-1} = N## for all ##a \in G##. Since this is an equality of sets, the usual way to prove it is to prove containment in both directions, namely ##aNa^{-1} \subseteq N## and ##N \subseteq aNa^{-1}##. The first containment is given (definition of normality), so you just need to prove the second one.

    Try thinking about my hint some more: ##aNa^{-1} \subseteq N## if and only if ##N \subseteq a^{-1}Na##
     
  8. Jul 31, 2014 #7

    micromass

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    To give a further hint. If you want to prove an inclusion such as ##N\subseteq aNa^{-1}##, you should always do the following; pick an arbitrary element ##x\in N## (so the only thing you know about ##x## is that it is in ##N##). You must proe that ##x\in aNa^{-1}##. In other words, you must find some ##y\in N## such that ##x=aya^{-1}##.

    (Yes, jbunniii, I know this is not the approach you had in mind, but I think this seems simpler)
     
  9. Jul 31, 2014 #8

    jbunniii

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    No problem, any proof is fine as long as it correct and (most importantly) Justabeginner understands it. I don't care if it's my proof or not. :smile:
     
  10. Jul 31, 2014 #9

    micromass

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    Your method is still important though, so be sure to give it after he finds the proof himself.
     
  11. Jul 31, 2014 #10
    Thank you both for the techniques.

    So suppose element x is in N, where x=aya^-1, then aya^-1 is contained in N. So, y is contained in a^-1Na and therefore if a^-1xa is contained in a^-1Na as well, x is contained in a^-1Na?
     
  12. Jul 31, 2014 #11

    jbunniii

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    How do you know that you can write ##x## in the form ##aya^{-1}##? It is true, but you haven't shown why.

    May I suggest instead starting as follows: ##x \in N##, and ##N## is a normal subgroup. Therefore, given any ##g \in G##, what can you say about ##gxg^{-1}##?
     
  13. Jul 31, 2014 #12
    If element x is in N, and N is a normal subgroup of G, given any element g in G, gxg^-1 must also be in G.

    Edit: gxg^-1 has to be in N too, because x is in N.
     
    Last edited: Jul 31, 2014
  14. Jul 31, 2014 #13

    micromass

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    Good. So what if we know define ##y## as ##axa^{-1}##?
     
  15. Jul 31, 2014 #14

    jbunniii

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    Small correction: "must also be in N."
     
  16. Jul 31, 2014 #15
    Now, if element y is contained in N, and y= axa^-1 then axa^-1 is also contained in N. If axa^-1 is in N and N is a subgroup of G, then axa^-1 is also in G, so a is in G.
     
  17. Jul 31, 2014 #16

    micromass

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    So we have the following:

    1) We know that ##y = axa^{-1}## by definition.
    2) We know that ##aNa^{-1} \subseteq N## since ##N## is normal by hypothesis.
    3) We know that ##x\in N##

    Can you use this to establish the following facts:

    4) ##x = aya^{-1}##
    5) ##y\in N##

    Finally, can you use those 5 facts to establish that ##x\in aNa^{-1}##?
     
  18. Jul 31, 2014 #17
    By multiplication, x=a^-1ya. Since x is in N, and x=a^-1ya, y is also in N. If y is in N, then a^-1xa is in N, and x is in aNa^-1.
     
    Last edited: Jul 31, 2014
  19. Jul 31, 2014 #18

    jbunniii

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    Shouldn't that be ##x = a^{-1}ya##?
     
  20. Jul 31, 2014 #19

    micromass

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    Yes, sorry for the confusion.
     
  21. Jul 31, 2014 #20
    I forgot to put them on the correct side when I multiplied.
    Is the flow of my statements right?
     
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