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Math Help - Cyclic group

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Cyclic group

    Let  G denote a group and let  g \in G . In each case show  G = < g > .

    a.  |G| = 12, g^4 \neq 1, g^6 \neq 1 .
    b.  |G| = 40, g^8 \neq 1, g^{20} \neq 1 .
    c.  |G| = 12, g^{30} \neq 1, g^{20} \neq 1, g^{12} \neq 1 .
    d. Generalize.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Let  G denote a group and let  g \in G . In each case show  G = < g > .

    a.  |G| = 12, g^4 \neq 1, g^6 \neq 1 .
    b.  |G| = 40, g^8 \neq 1, g^{20} \neq 1 .
    c.  |G| = 12{\color{red}0}, g^{30} \neq 1, g^{20} \neq 1, g^{12} \neq 1 .
    d. Generalize.
    The order of an element of a group always divides the order of the group (so in example c., |G| can't be 12, it must be 120).

    In example a., the order of g must be 1, 2, 3, 4, 6 or 12 (those being the divisors of 12. If g has order 1 then g is the identity and so g to any other power is also 1. If g has order 2 then g^4 = 1. If g has order 3 then g^6=1. So if g^4\ne1 and g^6\ne1 then the only remaining possibility is that g has order 12. Thus g generates the whole of G.

    The other two examples are similar. I'll leave you to think about the generalisation.
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