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All homomorphisms from Z_n to Z_m

  1. May 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Describe all group homomorphisms from [itex] \mathbb{Z}_n [/itex] to [itex] \mathbb{Z}_m [/itex].

    2. Relevant equations

    [itex] \mathbb{Z}_n = {[0],[1],\dots ,[n-1]} [/itex] with addition.

    A homomorphism is an operation preserving map, ie [itex] \phi (a\ast b)=\phi (a) \# \phi (b) [/itex].

    One especially important homomorphism property is that [itex] \phi (a^k) = \phi (a)^k [/itex].

    We can describe each homomorphism entirely by its action on any element that generate the group.

    3. The attempt at a solution

    I am pretty sure there are [itex] \text{gcd}(n,m) [/itex] homomorphisms from [itex] \mathbb{Z}_n [/itex] to [itex] \mathbb{Z}_m [/itex].

    Based on some examples I worked out, I believe the solution is:

    let [itex] [a] [/itex] be any element which generates [itex] \mathbb{Z}_n [/itex]

    [tex] \phi ([a]) = \frac{n}{\text{gcd}(n,m)}\cdot k [a] [/tex] where [itex] 0\le k < \text{gcd}(m,n) [/itex]
     
  2. jcsd
  3. May 21, 2013 #2

    I like Serena

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    You are correct, but your "description" can use improvement and I believe it contains a mistake.

    Let's simplify your generator a bit.
    Instead of picking [a], it's easier to pick [1] as a generator.

    Then candidates for the homomophisms are the ones defined by ϕ([1]).
    The restriction is that ϕ(n[1]) = n ϕ([1]) = 0 mod m.
    In other words m | n ϕ([1]).

    Now let g = gcd(m,n), then there are numbers a and b such that ##m=ga, n=gb, \gcd(a,b)=1##.

    Then:
    $$ga\ |\ gbϕ([1]) \\
    a\ |\ bϕ([1]) \\
    a\ |\ ϕ([1])$$
    Do you see why?

    So you're left with the question which and how many values of ϕ([1]) meet with the condition that ##\frac m {\gcd(m,n)}## divides ϕ([1]).
     
  4. May 21, 2013 #3
    I did have [1] in mind for [a], but it should work on any generator.

    I'm not sure what you are getting at with your last statement.

    You mentioned you thought there was a mistake, but you also said I was correct?
     
  5. May 21, 2013 #4

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    I believe the number of homomorphisms is indeed ##\gcd(m,n)##, but the formula you gave for ϕ([a]) is not entirely correct.
     
    Last edited: May 21, 2013
  6. May 21, 2013 #5
    So are you saying some of the maps specified by my formula are redundant, some are missing, or some are plain wrong?
     
  7. May 21, 2013 #6

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    Oh bollocks! I'm saying it should be:
    $$\phi([1]) = \frac m {\gcd(m,n)} k[1]$$
    with ##0 \le k < \gcd(m,n)##.

    Can you still smile? :shy:
    This should be fun.
     
  8. May 21, 2013 #7
    Oops. Thanks :)
     
  9. May 21, 2013 #8

    I like Serena

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    I guess of the choices offered, it fits "plain wrong". :biggrin:
     
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