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Math Help - Cyclic Groups

  1. #1
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    Cyclic Groups

    Hello,
    If A={1,a,a^2,a^3,a^5} is a set defined under a binary operation * can this be considered as a cyclic group.
    Since a^4 is not in the set A,I have concluded that,a*(a^3)=a^4 is not in the set A. And therefore * is not a binary operation,hence this is not a group.
    Is this conclution correct.Therefore in my mind a cyclic group must contain all the powers of the generator(without missing powers inbetween) i.e: The set A should be {1,a,a^2,a^3,a^4,a^5} in order to call a cyclic group under *.
    Are these things I have assumed correct?
    Please answer my question as quick as you can.

    Thank you.
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  2. #2
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    Going with the usual definition then yes you're missing a^4, but notice that I could define a cyclic group G= \{ 1,a^2,a^4,a^6\} where the exponents are taken modulo 8 then clearly G is a group although I'm missing some exponents (what I could do to get rid of this is take an isomorphism with C_4)
    Last edited by Jose27; December 30th 2009 at 08:54 PM.
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Hello,
    If A={1,a,a^2,a^3,a^5} is a set defined under a binary operation * can this be considered as a cyclic group.
    Since a^4 is not in the set A,I have concluded that,a*(a^3)=a^4 is not in the set A. And therefore * is not a binary operation,hence this is not a group.
    Is this conclution correct.Therefore in my mind a cyclic group must contain all the powers of the generator(without missing powers inbetween) i.e: The set A should be {1,a,a^2,a^3,a^4,a^5} in order to call a cyclic group under *.
    Are these things I have assumed correct?
    Please answer my question as quick as you can.

    Thank you.
    yes you can show this by contradiction, suppose A is a cyclic group, then by definitions of group, if a,b are members of group G, then so is a*b, however a*a^3 does not belong to A.
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  4. #4
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    Quote Originally Posted by Jose27 View Post
    Going with the usual definition then yes you're missing a^4, but notice that I could define a cyclic group G= \{ 1,a^2,a^4,a^6\} where the exponents are taken modulo 8 then clearly G is a group although I'm missing some exponents (what I could do to get rid of this is take an isomorphism with C_4)
    Dear Jose27,

    Can you please tell me what do you mean by taking an isomorphism. Since I have taken Group theory recently I hope you will explain it in a simple manner.

    Thank you.
    Last edited by Sudharaka; December 31st 2009 at 12:11 AM.
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Sudharaka View Post
    Hello,
    If A={1,a,a^2,a^3,a^5} is a set defined under a binary operation * can this be considered as a cyclic group.
    Since a^4 is not in the set A,I have concluded that,a*(a^3)=a^4 is not in the set A. And therefore * is not a binary operation,hence this is not a group.
    Is this conclution correct.Therefore in my mind a cyclic group must contain all the powers of the generator(without missing powers inbetween) i.e: The set A should be {1,a,a^2,a^3,a^4,a^5} in order to call a cyclic group under *.
    Are these things I have assumed correct?
    Please answer my question as quick as you can.

    Thank you.
    Surely defining * to be <a: a^3=1> will give you a cyclic group over this set. This will essentially say that a^5 = a^2, so we can just ignore it...

    I believe the answer is meant to be "no", but this set will form a group under this operation. There is nothing that I know of in the axioms that says every element in the set must be distinct with respect to the operation.

    I think.
    Last edited by Swlabr; December 31st 2009 at 06:29 AM.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Sudharaka View Post
    Dear Jose27,

    Can you please tell me what do you mean by taking an isomorphism. Since I have taken Group theory recently I hope you will explain it in a simple manner.

    Thank you.
    An isomorphism is a formal way of saying two group are "the same". For instance, taking the set G_1 = \{1, 2, 3, 4, 5, 6\} under multiplication modulo 7 will give us a group. It is cyclic, of order 6. However, this is "the same" as the group given by the set G_2 = \{0, 1, 2, 3, 4, 5\} under addition modulo 6, as both are cyclic of order 6. We say these groups are isomorphic. Formally, we map \theta: G_1 \rightarrow G_2, 3 \mapsto 1, 3^a \text{ mod } 7  = a \text{ mod} 6.

    Now, because this mapping preserves size (it is a bijection - the image and pre-image have the same cardinality) we are half way there. We also need it to preserve the operation. That is, we require (g\theta)(h\theta) = (gh)\theta. We call such mappings homomorphisms. This holds for our mapping \theta.

    A bijective homomorphism is called an isomorphism. It preserves both size and the operation.

    See here for wiki's definition. You will, undoubtedly, cover this very soon in your course.
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  7. #7
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    Dear Jose27,sym0110 and Swlabr,

    Thank you so mush for the time you had spent answering these questions. Now I have a clear idea about cyclic groups.Thanks a lot.
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