1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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##.

    $$ga\ |\ gbϕ([1]) \\
    a\ |\ bϕ([1]) \\
    a\ |\ ϕ([1])$$
    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

    I like Serena

    User Avatar
    Homework Helper

    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

    I like Serena

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    I guess of the choices offered, it fits "plain wrong". :biggrin:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: All homomorphisms from Z_n to Z_m
  1. Homomorphism from A4 (Replies: 1)