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Group commutativity problem

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Let G be a finite group whose order is not divisible by 3. Suppose (ab)3 = a3b3 ([itex]\forall[/itex]a,b[itex]\in[/itex]G)
    Prove G must be abelian.

    Known:
    G is a finite group
    o(G) not divisible by 3
    (ab)3 = a3b3 for all a,b[itex]\in[/itex]G

    2. Relevant equations
    θ:G → G s.t. θ(g) = g3 [itex]\forall[/itex]g [itex]\in[/itex] G
    Group axioms
    Homomorphism properties
    Lagrange's theorem
    Modular arithmetic?

    3. The attempt at a solution
    From the facts given above, it's obvious that θ is a homomorphism. Since o(G) is not divisible by 3, then by Lagrange's Theorem, no subgroup of g has order 3. In particular, o(g) ≠ 3 [itex]\forall[/itex]g [itex]\in[/itex] G. This implies that [itex]\forall[/itex]g [itex]\in[/itex] G a3 = e iff a = e, therefore ker θ = {e}.

    Then let a,b [itex]\in[/itex] G s.t. θ(a) = θ(b)
    => θ(ab-1) = θ(a)θ(b-1) = θ(a)θ(b)-1 = e
    => ab-1 [itex]\in[/itex] ker θ
    => ab-1 = e => a = b
    Thus θ is 1-1.
    Since G is finite, we can use the pigeonhole principle to prove θ is onto. Thus θ is a group automorphism. Since the set of all automorphisms of G is a group, then for any integer z, then θz is an automorphism of G.
    If I can prove that either functions i(g) = g-1 or d(g) = g2 are in <θ>, then I'm done, but I don't think that's true in general. Other than that I don't know how to proceed further, though.
     
    Last edited: Oct 9, 2011
  2. jcsd
  3. Oct 10, 2011 #2

    Deveno

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    Science Advisor

    this is an interesting problem!

    applying θ to aba-1, we get:

    ab3a-1 = a3b3a-3
    b3 = a2b3a-2
    b3a2 = a2b3

    but θ is an automorphism (bijective), so every element of G is uniquely a cube.

    that is, b3 = c for some unique c in G.

    so a2c = ca2, for all a,c in G.

    that is, a2b = ba2, for all a,b in G

    now because (ab)3 = a3b3:

    a(ba)2b = a(a2b2)b, so that

    (ba)2 = a2b2.

    starting from a2b = ba2, then, we get:

    a2b2 = ba2b
    (ba)2 = (ba)(ab)
    ba = ab.
     
  4. Oct 10, 2011 #3

    How'd you use θ to get ab3a-1 = a3b3a-3?
     
  5. Oct 10, 2011 #4

    Deveno

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    Science Advisor

    (aba-1)3 = (aba-1)(aba-1)(aba-1)

    = ab(aa-1)b(aa-1)ba-1 = ab(e)b(e)ba-1

    = ab3a-1,

    that is θ(aba-1) = ab3a-1.

    but θ is a homomorphism, θ(aba-1) = θ(a)θ(ba-1)

    = θ(a)θ(b)θ(a-1) = a3b3(a-1)3

    = a3b3a-3
     
  6. Oct 10, 2011 #5
    Ok, thanks
     
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