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Horrible problem about abelian groups

  1. Sep 12, 2011 #1
    G is a group and for all elements a,b in G,
    (ab)^i = (a^i)(b^i) holds for 3 consecutive positive integers.
    Show that G is abelian.

    I know how to prove that if (ab)^2 = (a^2)(b^2) then G is abelian. I was thinking that you could reduce the given equality integer by integer till 2 or something like that, but all I have been able to do so far is use the cancellation laws to get some gibberish. HAALP!
  2. jcsd
  3. Sep 12, 2011 #2
    OK, we can do this all formally, but it's best to do with an example. Let's assume that

    aaaabbbb = abababab

    aaaaabbbbb = ababababab

    aaaaaabbbbbb = abababababab

    Let's deal with the first two equalities first. Substitute abababab=aaaabbbb into aaaaabbbbb.
    Can you use this to prove that abababab = babababa ??
  4. Sep 12, 2011 #3
    baabaa black sheep
  5. Sep 12, 2011 #4
    starting with the 2nd equation, using the left and right cancellation laws on a and b you get
    aaaabbbb =babababa. Using this is with the first equation yields babababa=abababab
  6. Sep 12, 2011 #5
    OK, now do the same with the second and the third equation.
  7. Sep 12, 2011 #6
    3rd equation => aaaaabbbbb=bababababa
    plugging into the 2nd gives bababababa=ababababab
    then plug in babababa=abababab to cancel and get ab=ba
    thank you@!
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