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Math Help - Show Cyclic

  1. #1
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    Show Cyclic

    Let G be a group of order three. Show that G is cyclic. I feel like this should be a very simple problem, but I am not sure how to proceed.
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  2. #2
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    Re: Show Cyclic

    This is what I have so far:
    Let G be a group, g is an element of G. The order of g is the smallest positive integer n such that g^n=1. We know that n exists because G is finite. Furthermore, we are told the order is 3, so n=3. I have written g^n=1, and in this case, g^3=1, which implies that g=<1>. So, <1>={1,2,0}.
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  3. #3
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    Re: Show Cyclic

    this is what we're given:

    G = {e,a,b}.

    as yet, we know nothing about the order of a or b. but we can ask: which element of G is a2?

    we have 3 choices: e, a or b.

    if a2 = a, then a-1(a2) = a-1a, that is:

    a = e, a contradiction. so a2 must be e, or b. if a2 = e, then a is its own inverse. that means b must also be its own inverse (as there are no more elements left to be its inverse: b is not the identity, so b-1 ≠ e,

    and b-1 ≠ a, since a-1 is a).

    so far, so good. now we ask: if a and b are their own inverses, which element of G is ab?

    if ab = e, then:

    a(ab) = a
    a2b = a
    eb = a
    b = a, contradiction.

    if ab = a, then:

    a(ab) = a2
    a2b = e
    b = e, also a contradiction.

    if ab = b, then:

    (ab)b = b2 (since b = b-1, b2 = bb-1 = e)
    ab2 = e
    ae = e
    a = e, also a contradiction.

    so setting a2 = e leaves us with no possible choice for ab. since we have ruled out a2 = a, as well, that leaves only one possible choice:

    a2 = b, in which case:

    G = {e,a,b} = {e,a,a2} = <a>.

    (i want to point out that all we have done is shown that IF |G| = 3, then G is cyclic. we haven't actually shown that any groups of order 3 actually exist. however, one can easily verify that the set {e,a,b} with the operation * defined by:

    e*e = e
    e*a = a
    e*b = b

    a*e = a
    a*a = b
    a*b = e

    b*e = b
    b*a = e
    b*b = a

    actually defines a group with identity e. the hard part is verifying associativity, which is a chore (there are 9 possible products, so there are 27 possible triple products, which gives 27 associations to verify).
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  4. #4
    Senior Member jakncoke's Avatar
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    Re: Show Cyclic

    Let |G| = 3.  G = \{e,a,b\} . the order of an element divides the order of the group, since only 1 element of order 1 exists (e). |a| and |b| = 3. which means a,b are generators for G. thus G is cyclic.
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