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Showing a group is abelian

  1. Mar 6, 2006 #1

    StatusX

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    I have a couple questions involving showing a group with certain properties is abelian.

    1. For the first, I'm supposed to show that if some group G has the property that (ab)i=aibi for some three consecutive integers i and all a,b in G, then G must be abelian. Using (aba-1)i=abia-1=aibia-i, I've been able to show that a2b=ba2 for all a,b in G, but I can't get any farther.

    2. The second is similar. Given that a finite group G has order not divisible by 3, and for every a,b in G, (ab)3=a3b3, show G is abelian. By defining an automorphism on G by sending a to a3, I've been able to show every element has a unique cube root. Using this, I've shown a2b=ba2, as above. But now I'm stuck.

    Thanks in advance for any help.
     
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  3. Mar 6, 2006 #2

    AKG

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    Problem 1

    Your work doesn't look right. How do you get abia-1 = aibia-i?

    Let I = {i-2, i-1, i} be a set of three consecutive integers such that for all a, b in G, and all k in I, (ab)k = akbk

    Hints:

    1. Starting with (ab)i = aibi, deduce that (ba)i-1 = (ab)i-1

    2. Deduce, somehow, that (ba)i-2 = (ab)i-2
     
  4. Mar 6, 2006 #3

    AKG

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    Note, for problem 2, you don't need to show that x |-> x³ is an automorphism. Simply show that it is injective (it doesn't even need to be a homomorphism). Then, since G is finite, you know the map is surjective, and this is all you need to show that each element has a unique cube root.
     
  5. Mar 6, 2006 #4

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    Thanks for the help. (ba)i=b(ab)i-1a. It's so simple, and pops out at you if you write the product out, but I was going a completely different way with the problem and would never have thought of it. Both answers follow easily from it.

    And by the way, since conjugation by an element is a homomorphism (in fact, an automorphism), then (aba-1)i=abia-1.
     
  6. Mar 6, 2006 #5

    AKG

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    My issue wasn't with (aba-1)i = abia-1, it was with abia-1 = aibia-i. However, I now see where you got that:

    (aba-1)i = aibia-i by the special property given to us
    (aba-1)i = abia-1 because it's an easy fact for any group

    Using a similar idea for problem 2, you get:

    a³b³a-³ = ab³a-1
    a²b³ = b³a²

    so every square commutes with every cube. But since you can prove that every element is a cube (i.e. it has a cube root), you can say that every square commutes with every element. Not sure where to go from here.
     
  7. Mar 6, 2006 #6

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    Well, using a similar idea as the one you suggested for the first problem, you can show (ab)2=b2a2. Then you can pull a factor of b on the right side to the other side of a2, and its easy from there.
     
  8. Mar 6, 2006 #7

    AKG

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    When I first looked at the problem, I did get that (ab)² = b²a². However, I don't see where to go from there. In fact, what do you even mean by:

    Then you can pull a factor of b on the right side to the other side of a2, and its easy from there.
     
  9. Mar 6, 2006 #8

    AKG

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    Oh wait.

    (ab)³ = a³b³
    ababab = aaabbb
    baba = aabb
    (ba)² = a²b²

    Now this, in conjunction with the fact that the squares commute with everything, gives

    (ba)² = b²a²
    baba = bbaa
    ab = ba []

    Is that what you had in mind?
     
  10. Mar 6, 2006 #9

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    More or less. I actually specifically meant:

    (ab)2=b2a2
    abab=b(ba2)=b(a2b)=baab
    ab=ba
     
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