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Thread: 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 $\displaystyle G$ of order 6 must have at least one element $\displaystyle x$ of period 3.

    This is what I have so far.

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

    Maybe looking at an element $\displaystyle y$ in $\displaystyle G$ and $\displaystyle y$ is not in $\displaystyle <x>$, 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 $\displaystyle y$ idea is the right idea. Note that $\displaystyle \langle x\rangle\not=G.$ Therefore, there exists $\displaystyle y\in G$ such that $\displaystyle y\not\in\langle x\rangle.$ By hypothesis and assumption, $\displaystyle y\not =e,$ and $\displaystyle \langle y\rangle\not=G$ (or your group is cyclic, hence abelian). But, $\displaystyle |\langle y\rangle|$ divides $\displaystyle |G|,$ because it's a subgroup. Therefore,

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

    Now you'll need some clever argument as to why $\displaystyle |\langle y\rangle|\not=2.$ 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 $\displaystyle G/<x>$ and using the coset $\displaystyle y<x>$?

    I believe the factor group $\displaystyle G/<x>$ has order 3, since we are assuming $\displaystyle x$ 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 $\displaystyle G$ 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 $\displaystyle \langle x\rangle\trianglelefteq G?$ If so, then the OP'er's idea in post # 4, at the end, could work, if we can identify $\displaystyle G/\langle x\rangle$ with a subgroup of $\displaystyle G.$ 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 $\displaystyle \langle x\rangle\trianglelefteq G?$ If so, then the OP'er's idea in post # 4, at the end, could work, if we can identify $\displaystyle G/\langle x\rangle$ with a subgroup of $\displaystyle G.$ Can we do that?
    No, because it's not true: just take an arbitrary subgroup of $\displaystyle S_3$ 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 $\displaystyle G$ be such a group, so that $\displaystyle x^2=e$, for all $\displaystyle x\in G$.

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

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

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

    $\displaystyle ab=ba$

    and, since these elements were arbitrary, this implies that $\displaystyle G$ 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 $\displaystyle S_3$ 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 $\displaystyle G$ be such a group, so that $\displaystyle x^2=e$, for all $\displaystyle x\in G$.

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

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

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

    $\displaystyle ab=ba$

    and, since these elements were arbitrary, this implies that $\displaystyle G$ 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 $\displaystyle x$ has period 2 and that there exists a $\displaystyle y \in G$ and $\displaystyle y \not\in <x>$ (so we don't know the period of $\displaystyle y$). I don't think that the proof topspin1617 gave implies $\displaystyle y$ is commutative, thus not proving that $\displaystyle G$ 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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