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Are these lines logical?

  1. Jan 11, 2015 #1
    To make a very long story short, in a group theory problem I am working on, I need to prove this:

    ##A \lhd B \Rightarrow A'\neq A##,
    where ##A## and ##B## are finite and ##A'## is called the commutator subgroup:

    ##\begin{align}
    A' :&= [A, A] \\
    &= \langle [x, y] \mid x, y \in A \rangle \\
    &= \langle x^{-1}y^{-1}xy \rangle
    \end{align}##.
    Here are the lines I made out: Since ##A## is a normal subgroup, therefore it is commutative. For ##\forall x, \forall y \in A##,

    ##\begin{align}
    xy &= yx && (1)\\
    y^{-1}xy &= x && (2)\\
    y^{-1}xy &\in A && (3)\\
    x^{-1}y^{-1}xy &\notin A && (4)\\
    \langle x^{-1}y^{-1}xy \rangle &\neq A && (5)\\
    \langle [x, y] \rangle &\neq A && (6)\\
    [A, A] &\neq A && (7)\\
    A' &\neq A && (8)
    \end{align}##.​

    With this posting, I am very interested to know if the logic from line (3) to (4) and to (5) are valid, and if they are not, I would love to learn how they should be instead. Thank you for your time and help.
     
  2. jcsd
  3. Jan 11, 2015 #2

    haruspex

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    I can think of no basis at all for your step (3) to (4). Indeed, isn't the term on the left in (4) the identity element?
    Are you sure? A long time since I studied groups, but that doesn't sound familiar.
     
  4. Jan 11, 2015 #3
    Honestly I am so doubtful of this logic that I decided to put it up here as a question. Do you have any other suggestion then?
     
  5. Jan 11, 2015 #4

    haruspex

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    Why do you think what you are trying to prove is even true?
    Isn't the trivial group consisting of the identity {1} a normal subgroup of B? And is it not equal to its own commutator subgroup?
     
  6. Jan 11, 2015 #5
    What I am trying to imply is that since the definition of normal subgroup goes like this:
    ##N \lhd G \Leftrightarrow \forall n \in N, \forall g \in G, ng = gn##,​
    does this imply that ##n## is then abelian? Let me know and thanks.
     
  7. Jan 11, 2015 #6

    haruspex

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    No, that's not the definition of a normal subgroup. g-1ng is an element of N, but not necessarily n.
    http://en.wikipedia.org/wiki/Normal_subgroup
     
  8. Jan 11, 2015 #7
    Let me type in the whole question so that you can see it as a big picture, give me about 5 minutes, if you will. Thanks.
     
  9. Jan 11, 2015 #8
    I typed in the original question couple of days on MSE here, but I did not get any help perhaps because I typed in the question way TOO LONG, therefore nobody gets interested and therefore it is lost forever, and there is no way to resuscitate it back from dead. I would love to get any idea from you. Thanks.
     
  10. Jan 11, 2015 #9

    haruspex

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    Ok. In this thread you omitted the key constraint that A is not simply {1}.
    Can you show that if Hi/Hi-1 is abelian then Hi-1 = Hi'?
     
  11. Jan 11, 2015 #10
    Is this the direction you are hinting me to pursue: If I can prove that ##H_{i-1} = H'_i##, then it implies ##H_i \neq H'_i##?
     
  12. Jan 11, 2015 #11

    haruspex

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    No, that would be trivial. ##H_{i-1} \neq H_i## by definition, so if ##H_{i-1} = H'_i##, then ##H_i \neq H'_i##.
     
  13. Jan 11, 2015 #12
    Ok, here is what I understand of factor group: If ##b \in H_i## then the factor group ##H_i / H_{i-1} = \{ bH_{i-1} \}##, am I correct? And then since ##H_i / H_{i-1}## is abelian, therefore ##bH_{i-1} = H_{i-1}b##? Am I correct?
     
  14. Jan 11, 2015 #13

    haruspex

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  15. Jan 11, 2015 #14
    Thanks! I think I am about to get it now, I will keep working on it. Thanks again and again.
     
  16. Jan 11, 2015 #15
    Are you still there? I am continuing the conversation, suppose that ##N## is normal subgroup of ##G## and ##G/H## is abelian:

    ##\begin{align}
    \forall aH,bH &\in G/H \\
    aHbH &= bHaH \qquad \qquad &&(1) \\
    abH &= baH &&(2) \\
    ab(ba)^{−1} &\in H &&(3a) \\
    ab(ba)^{−1} &=H &&(3b) \\
    \ldots & \ldots \\
    aba^{−1}b^{−1} &= H &&(4)\\
    [a, b] &= H &&(5)\\
    G' &= H &&(6) \\
    \end{align} ##
    My question is: Which one is correct, the (3a) or (3b)? As you advise, my final goal is ##G' = H##. Please advise and thanks again.

     
  17. Jan 11, 2015 #16

    haruspex

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    I assume you mean H, not N.
    Clearly 3(b) is not right, but maybe you meant ##<ab(ba)^{−1}> =H##.
    My reading of the link I posted is that you can only say ##<ab(ba)^{−1}> \leq H##.
    But maybe that's enough to get to the result you are after.
     
  18. Jan 11, 2015 #17
    Thank you for your time and help.
     
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