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Math Help - Group theory

  1. #1
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    Group theory

    Hi,

    G={1,5,25,125} is a cyclic group under modulo 625. Is this correct?

    I had a course on Group theory recently. Based on the answer to the above question I want to know whether I had understand the concepts well.

    Thank you.
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  2. #2
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    Quote Originally Posted by Sudharaka View Post
    Hi,

    G={1,5,25,125} is a cyclic group under modulo 625. Is this correct?

    I had a course on Group theory recently. Based on the answer to the above question I want to know whether I had understand the concepts well.

    Thank you.
    First you have to show that G is a group under the operation specified, which I will assume is multiplication modulo 625.

    Then you have to show that there is an element that generates the entire group.

    G fails to be a group as it is not closed under multiplication modulo 625 since:

    125^2 \equiv 0 \text{ (mod 625)} \not\in G

    CB
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  3. #3
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    Thank you very much for your reply. Yes you are correct the group was'nt closed under multiplication modulo 625. Now then suppose,

    G={1,5,25,125} is a cyclic group under modulo 626. Is this correct?
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    First you have to show that G is a group under the operation specified, which I will assume is multiplication modulo 625.

    Then you have to show that there is an element that generates the entire group.

    G fails to be a group as it is not closed under multiplication modulo 625 since:

    125^2 \equiv 0 \text{ (mod 625)} \not\in G

    CB

    125^2=(5^3)^2=5^6=5^2\!\!\!\!\pmod{625=5^4}\neq 0\!\!\!\!\pmod{625} , but the explanantion's still correct, of course: 25^2=5\cdot 125=0\!\!\!\!\pmod{625}\notin \{1,5,25,125\} and then this set isn't closed under multiplcation modulo 625.

    Tonio
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  5. #5
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    Quote Originally Posted by Sudharaka View Post
    Thank you very much for your reply. Yes you are correct the group was'nt closed under multiplication modulo 625. Now then suppose,

    G={1,5,25,125} is a cyclic group under modulo 626. Is this correct?


    Again no: 5^5=3125=621=-5\!\!\!\!\pmod{626}\notin G ...

    If you want to make sure to get a MULTIPLICATIVE subgroup modulo n, for some integer n, take any INVERTIBLE element x\in\mathbb{Z}_n (i.e., any element x\in\mathbb{Z}_n\,\,\,s.t.\,\,\,(x,n)=1 and then take G:=<x>=\{1,x,x^2,\ldots,x^{r-1}\}\,,\,\,with\,\,\,r=ord_n(x) .

    Tonio
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  6. #6
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    Quote Originally Posted by tonio View Post



    125^2=(5^3)^2=5^6=5^2\!\!\!\!\pmod{625=5^4}\neq 0\!\!\!\!\pmod{625} , but the explanantion's still correct, of course: 25^2=5\cdot 125=0\!\!\!\!\pmod{625}\notin \{1,5,25,125\} and then this set isn't closed under multiplcation modulo 625.

    Tonio
    According to my calculator:

    125^2=5^6=5^4 \times 25=625 \times 25\equiv 0 \text{ (mod 625)}


    CB
    Last edited by CaptainBlack; January 2nd 2010 at 05:04 AM.
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  7. #7
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    Thanks so much to all of you. Now for the last time,

    G={1,5,25,125} under multiplication modulo 624 is a cyclic group. Is'nt it?
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  8. #8
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    Quote Originally Posted by CaptainBlack View Post
    According to my calculator:

    125^2=5^6=5^4 \times 25=625 \times 25\equiv 0 \text{ (mod 625)}


    Of course you're right and even more: the above's exactly the calculation I wrote, but for some reason hidden deep in my brains and kept jealously by my 3 working neurons, I did that calculation ADDITIVELY modulo 625...a shame.

    Tonio


    CB
    .
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  9. #9
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    Quote Originally Posted by Sudharaka View Post
    Thanks so much to all of you. Now for the last time,

    G={1,5,25,125} under multiplication modulo 624 is a cyclic group. Is'nt it?

    This time yes, since (5,624)=1\,\,\,and\,\,\,5^4=1\!\!\!\!\pmod{624} , so that G=<5>\!\!\!\!\pmod{624} , a group of order 4 (since 625=1\!\!\!\!\pmod{624})

    Tonio
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  10. #10
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    Thank you tonio. The problem is cleared. I appreciate the help given by all of you. It's nice to know that such people like you are out there willing to help at any time.
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