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Math Help - Representatives of classes theorem

  1. #1
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    Representatives of classes theorem

    Suppose {a1, a2,......, am} is a complete set of representatives for Z/mZ

    Show if gcd(b, m) > 1, then {ba1, ba2, ......, bam} is not a complete set of representatives.

    So far I have
    [a]m ≡ [n]m
    [ba]m ≡ [bn]m ≠ [n]m

    not sure if this constitutes much of a proof though.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Re: Representatives of classes theorem

    Quote Originally Posted by scruz10 View Post
    Suppose {a1, a2,......, am} is a complete set of representatives for Z/mZ

    Show if gcd(b, m) > 1, then {ba1, ba2, ......, bam} is not a complete set of representatives.

    So far I have
    [a]m ≡ [n]m
    [ba]m ≡ [bn]m ≠ [n]m

    not sure if this constitutes much of a proof though.
    This basically comes down to the fact that a number 0\leq i<m generates \mathbb{Z}/m\mathbb{Z} if and only if gcd(m, i)=1. (Hint: use the Euclidean algorithm).

    Can you see how your result follows from this?
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  3. #3
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    Re: Representatives of classes theorem

    That's pretty much what I have to prove. I already proved the first part where gcd(b,m)=1. I just don't know how to prove it is not a full set if it is greater than 11

    Assume
    [ba]m ≡ [bn]m ≡ [n]m
    bn - n = mx
    n(b - 1) = mx
    b - 1 = mx/n
    b - mx/n = 1
    Therefore (b,m) = 1
    Contradiction!
    only thing is I can't tell if x/n is an integer.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Re: Representatives of classes theorem

    Quote Originally Posted by scruz10 View Post
    That's pretty much what I have to prove. I already proved the first part where gcd(b,m)=1. I just don't know how to prove it is not a full set if it is greater than 11

    Assume
    [ba]m ≡ [bn]m ≡ [n]m
    bn - n = mx
    n(b - 1) = mx
    b - 1 = mx/n
    b - mx/n = 1
    Therefore (b,m) = 1
    Contradiction!
    only thing is I can't tell if x/n is an integer.
    Assume it is a full set, then you have that ba_i + cm=1 for some i and some c\in \mathbb{Z} (by the definition of modulus), so...
    Last edited by Swlabr; July 7th 2011 at 01:08 AM.
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  5. #5
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    Re: Representatives of classes theorem

    what you want to show is that the mapping [a_j]\to [ba_j] (j = 1,....,m) is not 1-1 if gcd(b,m) ≠ 1.

    suppose that [0] = [m] = [a_s], and let gcd(b,m) = d > 1.

    since d divides m, there is some 1 ≤ u < m with du = m (we are using the fact that d > 1 here).

    note that [u] ≠ [m]. we'll use this later.

    since d divides b, there is some integer v with dv = b. let [a_t] = [u]

    (we can do this because we have a complete set of representatives, so u is in one of them).

    clearly, [ba_s] = [b][a_s] = [b][0] = [0b] = [0].

    (if a_s is a multiple of m, surely ba_s is as well).

    however, [ba_t] = [b][a_t] = [b][u] = [bu] = [(dv)u] = [v(du)]

     = [vm] = [v][m] = [v][0] = [0v] = [0].

    so [ba_s] = [ba_t] but [a_s] \neq [a_t].

    so since the map [a_j]\to [ba_j] is not 1-1, it cannot be onto (since our set of representatives is finite),

    so \{ba_1,ba_2,\dots ,ba_m\} is not a complete set of representatives.

    (i'm not usually so fussy about the brackets, but since the aj are presumably numbers, i use the brackets to indicate the entire equivalence class. hopefully you understand already that multiplication is well-defined on the equivalence classes, that is, that [a][b] = [ab], no matter "which" representatives we choose).
    Last edited by Deveno; July 6th 2011 at 06:50 AM.
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  6. #6
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    Re: Representatives of classes theorem

    Quote Originally Posted by Deveno View Post
    what you want to show is that the mapping [a_j]\to [ba_j] (j = 1,....,m) is not 1-1 if gcd(b,m) ≠ 1.

    suppose that [0] = [m] = [a_s], and let gcd(b,m) = d > 1.

    since d divides m, there is some 1 ≤ u < m with du = m (we are using the fact that d > 1 here).

    note that [u] ≠ [m]. we'll use this later.

    since d divides b, there is some integer v with dv = b. let [a_t] = [u]

    (we can do this because we have a complete set of representatives, so u is in one of them).

    clearly, [ba_s] = [b][a_s] = [b][0] = [0b] = [0].

    (if a_s is a multiple of m, surely ba_s is as well).

    however, [ba_t] = [b][a_t] = [b][u] = [bu] = [(dv)u] = [v(du)]

     = [vm] = [v][m] = [v][0] = [0v] = [0].

    so [ba_s] = [ba_t] but [a_s] \neq [a_t].

    so since the map [a_j]\to [ba_j] is not 1-1, it cannot be onto (since our set of representatives is finite),

    so \{ba_1,ba_2,\dots ,ba_m\} is not a complete set of representatives.

    (i'm not usually so fussy about the brackets, but since the aj are presumably numbers, i use the brackets to indicate the entire equivalence class. hopefully you understand already that multiplication is well-defined on the equivalence classes, that is, that [a][b] = [ab], no matter "which" representatives we choose).
    Seriously awesome....thanks
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