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|ab|=|ba| in a group

  1. Feb 2, 2004 #1
    I have to show that in a group G |ab|=|ba| for all a,b in G.

    I don't have a clue how to proceed. I assume |ab|=n. I feel that using associativity, identity and inverse should be important. But I don't see how these tools will get me to |ba|=n.
    I've considered proof by contradiciton, but that doesn't seem to be useful here.

    Can someone please give me a hint?
     
  2. jcsd
  3. Feb 2, 2004 #2

    Hurkyl

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    Well, I guess my first question is what you mean by |x|.
     
  4. Feb 2, 2004 #3
    Sorry, I figured |ab| would be understood to mean the order of ab; that is, the least positive number n such that (ab)^n=e. I'm new in the study of algebra, so I do not know what notation is most widely recognized, or in what contexts.
     
  5. Feb 3, 2004 #4

    Hurkyl

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    I thought that might be what you mean, but I wasn't sure; I've only seen that notation used for the order of a group, not the individual elements.


    Anyways, try this:

    Start with the equation [itex](ab)^n=e[/itex]. What can you do to this equation to turn the left hand side into [itex](ba)^n[/itex]?

    Oh, and don't forget this fun bit of number theory: [itex]a | b \wedge b | a \implies a = b[/itex] (if both a and b are positive integers)
     
  6. Feb 3, 2004 #5
    here's one way i think it could work:
    prove [tex]a\left( ba\right) ^{n}a^{-1}=\left( ab\right) ^{n}[/tex] (by induction, eg).

    then [tex]\left( ba\right) ^{\left| ba\right| }=e\rightarrow \left( ab\right) ^{\left| ba\right| }=a\left( ba\right) ^{\left| ba\right| }a^{-1}=e\rightarrow \left| ab\right| |\left| ba\right| [/tex].

    then prove [tex]\left( ba\right) ^{n}=a^{-1}\left( ab\right) ^{n}a[/tex] to get that [tex]\left| ba\right| |\left| ab\right| [/tex].
     
  7. Feb 3, 2004 #6

    NateTG

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    Phoenixtoth: Your first line of LaTeX is not generally true. Although for your argument you could go with [tex]a(e)a^{-1}=e[/tex] which is obviously true.
    Jupiter:
    Perhaps you should look at some simple cases:

    If [tex]|ab|=1[/tex] you should have little trouble.

    Once you've figured that one out, try [tex]|ab|=2[/tex]. You should be able to work it out from there.

    I think you'll have to handle the [tex]|ab|=\infty[/tex] case seperately, and that it will involve some type of proof by contradiction.
     
  8. Feb 3, 2004 #7
    nateTG,
    if it's not true, can you find a counter example?

    here's my proof by induction. can you spot an error?
    [tex]a\left( ba\right) ^{n}a^{-1}=\left( ab\right) ^{n}[/tex] is clear for n=0 if empty products are defined to equal e.

    assume now that [tex]a\left( ba\right) ^{n}a^{-1}=\left( ab\right) ^{n}[/tex] for n>0. multiply the left hand side by [tex]a^{-1}[/tex] and the right hand side by [tex]a[/tex] to get this:
    [tex]\left( ba\right) ^{n}=a^{-1}\left( ab\right) ^{n}a[/tex]. then
    [tex]a\left( ba\right) ^{n+1}a^{-1}=a\left( ba\right) ^{n}\left( ba\right) a^{-1}=a\left( a^{-1}\left( ab\right) ^{n}a\right) \left( ba\right) a^{-1}=\left( ab\right) ^{n+1}[/tex].
     
  9. Feb 3, 2004 #8

    NateTG

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    Sorry. Dyslexia strikes again. I thought you had:
    [tex]a(ab)^na^{-1}=(ab)^n[/tex] which is not generally true.
     
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