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Math Help - Group on Five Elements

  1. #1
    Member Haven's Avatar
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    Group on Five Elements

    I am having troubles proving this:
    Prove that any group on five elements is cyclic.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Haven View Post
    I am having troubles proving this:
    Prove that any group on five elements is cyclic.
    Do you know Lagrange's theorem?

    Lemma: If a set has no proper non-trivial subgroups, then it's cyclic.

    Proof: Let G meet the criterion. Let g\in G-\{e\} be arbitrary. We must then have that \left\langle g\right\rangle is a subgroup of G that is non-trivial. Thus, by assumption we must have that \left\langle g\right\rangle is improper, thus equal to G. The conclusion follows. \blacksquare

    Now, using this lemma and Lagrange's theorem we see that any subgroup of G where \left|G\right|=p for some prime p must be a divisor of p. Thus, any subgroup must be of order 1,p. Therefore, the lemma implies that G is cyclic for any prime-ordered group.

    Your example is just a corollary.
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  3. #3
    Member Haven's Avatar
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    I'm supposed to prove the statement by arguing that all the non-identity elements of G have order 5.

    So far I have
    Let x \in G such that x \neq e.
    so |x| \in \{2,3,4,5\}
    if |x| = 5, then we are done, since G = <x>.

    Case 1
    |x| = 4
    Then <x> = \{e,x,x^2,x^3\} and G = \{e,x,x,x^3,y\} where y \notin <x>.
    If xy = x,y \rightarrow x,y=e
    If xy = x^2,x^3 \rightarrow y = x,x^2 respectively
    If xy = e \rightarrow y=x^3

    Since they are all contradictory, x cannot have order 4. So no element in G can have order 4.

    But I can't figure out the other cases
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Haven View Post
    I'm supposed to prove the statement by arguing that all the non-identity elements of G have order 5.

    So far I have
    Let x \in G such that x \neq e.
    so |x| \in \{2,3,4,5\}
    if |x| = 5, then we are done, since G = <x>.

    Case 1
    |x| = 4
    Then <x> = \{e,x,x^2,x^3\} and G = \{e,x,x,x^3,y\} where y \notin <x>.
    If xy = x,y \rightarrow x,y=e
    If xy = x^2,x^3 \rightarrow y = x,x^2 respectively
    If xy = e \rightarrow y=x^3

    Since they are all contradictory, x cannot have order 4. So no element in G can have order 4.

    But I can't figure out the other cases
    Jeez. You basically have to construct the group table....good luck!
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  5. #5
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    Quote Originally Posted by Haven View Post
    I am having troubles proving this:
    Prove that any group on five elements is cyclic.
    Fascinating. I was looking up the Taoist 5 Elements and I found your thread about mathematical 5 elements.

    Not being a mathematician, I don't get it. Would you mind giving me the basic premise of the mathematical 5 elements concept?

    In Taoism, any five cyclical system is represented by wood, fire, earth, metal, and water. How each supports each other is obvious - back down to water, the fifth element supporting wood, the first element all over again. This characterization is supposed to be a perfect analogy for anything set of five that are mutually supportive. I found a fuller description of the 5 elements
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  6. #6
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    Quote Originally Posted by Haven View Post
    I'm supposed to prove the statement by arguing that all the non-identity elements of G have order 5.
    But that was exactly was Drexel28 did! Drexel28 proved that if g\in G and g\ne e, then \langle g\rangle=G which is to say that g has order 5.
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