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Math Help - help me clean up a grp thry proof

  1. #1
    Newbie TheSicklyElf's Avatar
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    Post help me clean up a grp thry proof

    My proof of this problem looks clunky...suggestions for cleanup or different approaches are welcome. Sorry, I really need to get laTeX, this looks ridiculous without it.

    From 2nd ed. of I.N. Herstein's Topics in Algebra:
    Let G be a finite group whose order is not divisible by 3. Suppose that (ab)^3=a^3b^3 for all a,b in G. Prove that G must be abelian.

    ----
    Denote the order of G by o(G) and G's identity element by e.

    We first show that there can be no a in G other than e for which a^3=e. Since if G is a finite group and a is in G, then a^o(G)=e (Corollary 2, page 43, a consequence of the definition of the order of a group element). Then a^o(G)=e=a^3n for any integer n>0. But if 3 does not divide o(G) then o(G)=3m+r, where 3 does not divide m>0 and 0<r<3, m,r integers. Thus a^o(G)=a^(3m+r)=a^(3m)a^r=a^r, one of the two distinct powers of a which do not equal a^0=a^3=e, contradicting a^o(G)=e.

    Now for all a,b in G, (ab)^3=ababab=a^3b^3. Cancellation shows baba=a^2b^2. The equation must be true when a and b are not each other's inverse and when a does not equal b does not equal e (that is, a, b and e are distinct from each other). But any association on baba (namely, (b)(aba), (ba)(ba), and (bab)(a)) forces a=b. The equation can only be true for all a,b in G if all a,b in G can be commuted. Thus G is abelian.



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  2. #2
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    Beautiful problem! (It was posted here before).

    Define the homomorphism:
    \phi : G\mapsto G as \phi (x) = x^3.
    Notice that,
    \ker \phi = \{ 1 \}.
    So, \phi is 1-to-1 (it is not divisible by 3).
    By Pigeonhole Principle \phi is surjective.

    Now consider,
    a^3b^3a^{-3} = (aba^{-1})^3= ab^3a^{-1}
    So,
    a^2b^3 = b^3a^2
    Since \phi is surjective b^3 can represent any element in G so,
    a^2x=xa^2 \mbox{ for all }x\in G.
    Hence,
    a^2 \in Z(G) the center of G (because it commutes with everything).

    Now,
    (ab)^3=a^3b^3
    ababab=a^3b^3
    baba=a^2b^2
    baba=b^2a^2 ---> since a^2 \in Z(G).
    ab=ba
    Q.E.D.

    ----------
    Try doing the problem after that in Herstein's book
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  3. #3
    Newbie TheSicklyElf's Avatar
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    Sweet! That's really quite a nice proof. lol there's like, 2 lines that show what I took the whole first paragraph of my proof to show. There's just one part I don't follow, which is how phi is one-to-one. Elaborate a bit? I tried to figure it out but I feel I'm wasting time :P
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  4. #4
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    Quote Originally Posted by TheSicklyElf View Post
    Sweet! That's really quite a nice proof. lol there's like, 2 lines that show what I took the whole first paragraph of my proof to show. There's just one part I don't follow, which is how phi is one-to-one. Elaborate a bit? I tried to figure it out but I feel I'm wasting time :P
    Are you familar with the rule that if the kernel is only the identity element then the map is one-to-one?

    Say,
    \phi(x) = 1
    Then,
    x^3 = 1
    That means,
    x=1 because otherwise if x\not = 1 then the order of x is three and hence \left< x \right> is a cyclic subgroup of order 3. But by Lagrange's theorem then 3 is a divisor of the order of this group ---> a contradiction!
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  5. #5
    Newbie TheSicklyElf's Avatar
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    OK I finally get it; if phi's kernel is 1 and phi(x)=phi(y) you have phi(x)[phi(y)]^(-1)=1 and following that logic, x=y.
    Yeah Herstein has a lot of good problems. Thanks for sharing your proof, I definitely learned from it.
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