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Math Help - Group Theory

  1. #1
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    Group Theory

    Hi anyone knows how to solve this?

    i) Consider Z*15, the group of invertible element of Z15 under the operation of multiplication (mod 15). What is the size of Z*15?
    ii) Consider Z*6, the group of invertible element of Z6 under the operation of multiplication (mod 6). What is the size of Z*6?

    Appreciate if anyone can help me (:
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  2. #2
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    Re: Group Theory

    Recall that the multiplicative inverse modulo n is found using the Bézout's identity. Also recall that the number of positive integers less than or equal to n that are relatively prime to n is given by the Euler's totient function.

    For more, see Multiplicative group of integers modulo n.
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  3. #3
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    Re: Group Theory

    The key thing to remember is that " x has multiplicative inverse in \mathbb{Z}_n^* if and only if x and n are relatively prime."
    (One way is straightforward to prove, the other way is from Bezout's identity.)

    The details are found in emakarov's post and its links.
    Last edited by johnsomeone; October 10th 2012 at 06:07 AM.
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  4. #4
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    Re: Group Theory

    Guys, i still dun get it after reading the links will it be possible if u can enlighten me
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  5. #5
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    Re: Group Theory

    Do you know what the definition is of an invertible element in \mathbb{Z}_n? If yes, then try to apply it, one element at a time, to \mathbb{Z}_{15}. In other words, go through the non-zero elements of \mathbb{Z}_{15}, and for each, either find the inverse, or declare that it's not invertible. Then count the number of elements that have an inverse.

    I'll do \mathbb{Z}_{6} for you:

    In \mathbb{Z}_{6}, [1] [1] = [1], so [1] is invertible, and its inverse is [1].

    In \mathbb{Z}_{6}, [2] [3] = [0], so [2] is not invertible (if [2] [a] = [1], then [2] [a] [3]= [1] [3] = [3], so [0] = [3]. Impossible.)

    In \mathbb{Z}_{6}, [3] [2] = [0], so [3] is not invertible (same reasoning as with [2].)

    In \mathbb{Z}_{6}, [4] [3] = [12] = [0], so [4] is not invertible (same reasoning as with [2].)

    In \mathbb{Z}_{6}, [5] [5] = [25] = [24+1] = [(6)(4)+1] = [1], so [5] is invertible, with inverse [5].

    Thus \mathbb{Z}_{6}^* = \{ [1], [5] \}, and so |\mathbb{Z}_{6}^*| = 2.

    Here are a few from \mathbb{Z}_{15}:

    In \mathbb{Z}_{15}, [1] [1] = [1], so [1] is invertible, and its inverse is [1].

    In \mathbb{Z}_{15}, [2] [8] = [16] = [1], so [2] is invertible, and its inverse is [8].

    In \mathbb{Z}_{15}, [3] [5] = [15] = [0], so [3] is not invertible.

    In \mathbb{Z}_{15}, [4] [4] = [16] = [1], so [4] is invertible, and its inverse is [4].

    ... (try to fill these in) ...

    In \mathbb{Z}_{15}, [13] [7] = [91] = [(15)(6)+1] = [1], so [13] is invertible, and its inverse is [7].

    In \mathbb{Z}_{15}, [14] [14] = [196] = [195+1] = [(15)(13)+1] = [1], so [14] is invertible, and its inverse is [14].
    Last edited by johnsomeone; October 14th 2012 at 09:26 AM.
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  6. #6
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    Re: Group Theory

    Thanks! I worked out.
    I got |Z*15| = {[1],[2],[4],[7],[8],[13],[14]}

    (:
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  7. #7
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    Re: Group Theory

    Miss out 11.
    I got |Z*15| = {[1],[2],[4],[7],[8],[11],[13],[14]}

    Thanks you very much!
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  8. #8
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    Re: Group Theory

    If the problem makes sense to you now, you might want to go back and reread emakarov's post and its links. That now might make more sense.
    Last edited by johnsomeone; October 16th 2012 at 01:43 PM.
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