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Math Help - Non-abelian group and periods of elements

  1. #1
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    Non-abelian group and periods of elements

    Prove that a non-abelian group G of order 6 must have at least one element x of period 3.

    This is what I have so far.

    Let x in G be a non-identity element. x can have period 2, 3, or 6.
    Period of x cannot be 6 because that make G cyclic, thus abelian.
    If x has period 3, we are done.
    If x has period 2, ... (stuck here)

    Maybe looking at an element y in G and y is not in <x>, but not sure what to do with it.

    Thanks in advance for any help.
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  2. #2
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    Have you done the Sylow theorems yet? Seems like the Corollary to # 1 there might be useful.
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  3. #3
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    Wait, here you go:

    I think your y idea is the right idea. Note that \langle x\rangle\not=G. Therefore, there exists y\in G such that y\not\in\langle x\rangle. By hypothesis and assumption, y\not =e, and \langle y\rangle\not=G (or your group is cyclic, hence abelian). But, |\langle y\rangle| divides |G|, because it's a subgroup. Therefore,

    |\langle y\rangle|\in\{2,3\}.

    Now you'll need some clever argument as to why |\langle y\rangle|\not=2. Thinking...
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  4. #4
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    Sorry I have not heard of Sylow Theorems yet.

    Is it possible to use cosets somehow? Using the factor group G/<x> and using the coset y<x>?

    I believe the factor group G/<x> has order 3, since we are assuming x has period 2.
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  5. #5
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    Well, if there are no elements of order 3, then the five non-identity elements must all have order 2 (because as you noted, an element of order 6 would make the group cyclic, hence abelian).

    Can you prove that any group whose non-identity elements all have order 2 must be abelian? This would of course give an immediate contradiction to G not being abelian.

    EDIT: Sorry, using the word "order" where you use "period".
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  6. #6
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    Reply to topspin1617:

    That is exactly the lines of thinking I was contemplating. But I'm not sure how to do it.
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  7. #7
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    Can we guarantee that \langle x\rangle\trianglelefteq G? If so, then the OP'er's idea in post # 4, at the end, could work, if we can identify G/\langle x\rangle with a subgroup of G. Can we do that?
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  8. #8
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    Quote Originally Posted by Ackbeet View Post
    Can we guarantee that \langle x\rangle\trianglelefteq G? If so, then the OP'er's idea in post # 4, at the end, could work, if we can identify G/\langle x\rangle with a subgroup of G. Can we do that?
    No, because it's not true: just take an arbitrary subgroup of S_3 of order 2, you can show that it isn't normal.

    But the fact that a group of exponent 2 is abelian is nearly trivial. Let G be such a group, so that x^2=e, for all x\in G.

    Suppose a,b\in G are arbitrary. Consider the relation

    (ab)^2=e
    abab=e
    ababb^{-1}=eb^{-1}
    aba=b^{-1}
    abaa^{-1}=b^{-1}a^{-1}
    ab=b^{-1}a^{-1}.

    But if an element has order 1 or 2, it is clearly its own inverse. So the last relation simplifies to

    ab=ba

    and, since these elements were arbitrary, this implies that G is abelian.
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  9. #9
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    Quote Originally Posted by topspin1617 View Post
    No, because it's not true: just take an arbitrary subgroup of S_3 of order 2, you can show that it isn't normal.

    But the fact that a group of exponent 2 is abelian is nearly trivial. Let G be such a group, so that x^2=e, for all x\in G.

    Suppose a,b\in G are arbitrary. Consider the relation

    (ab)^2=e
    abab=e
    ababb^{-1}=eb^{-1}
    aba=b^{-1}
    abaa^{-1}=b^{-1}a^{-1}
    ab=b^{-1}a^{-1}.

    But if an element has order 1 or 2, it is clearly its own inverse. So the last relation simplifies to

    ab=ba

    and, since these elements were arbitrary, this implies that G is abelian.
    Ah, very nice. Note to Zalren: this post # 8 pretty much solves your problem, I believe.
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  10. #10
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    So I understand what topspin1617 is saying, but does it apply? I am working under the assumption that x has period 2 and that there exists a y \in G and y \not\in <x> (so we don't know the period of y). I don't think that the proof topspin1617 gave implies y is commutative, thus not proving that G is abelian and creating a contradiction.
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  11. #11
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    What topspin is saying is that suppose there is no element of order 3. Then all five non-identity elements are of order 2. Then you do the computation topspin did, and you show that any two elements commute, contrary to the non-abelian assumption.
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  12. #12
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    Ah I see, thank you!
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  13. #13
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    Well, obviously topspin did the majority of the work here, but you're certainly welcome for my contribution.
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