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Prove only group homomorphism between Z5 and Z7 is the trivial one.

  1. Aug 22, 2013 #1
    1. The problem statement, all variables and given/known data.

    Prove that the only homomorphism between Z5 and Z7 is ψ(x)=0 (the trivial homomorphism).

    3. The attempt at a solution.

    I wanted to check if my solution is correct, so here it goes:

    Any element x in Z5 belongs to the set {0,1,2,3,4}

    So, I trivially start by saying that ψ(1)=ψ(1) and then, using the fact that ψ(x+y)=ψ(x)+ψ(y), I get:
    ψ(2)=ψ(1+1)=ψ(1)+ψ(1)=2ψ(1)
    .
    .
    .
    ψ(5)=ψ(1+1+1+1+1)=ψ(1)+ψ(1)+ψ(1)+ψ(1)+ψ(1)=5ψ(1)

    But 5=0 (mod 5) and, as ψ is a homomorphism, it must be 0= ψ(0)=ψ(5).
    By transitivity, 0=5ψ(1) (mod 7) but 5≠0 (mod 7), then ψ(1)=0. But ψ(x)=xψ(1)=0 for all x in Z5, hence Hom(Z5,Z7)=0.
     
  2. jcsd
  3. Aug 22, 2013 #2

    Office_Shredder

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    I think this last line needs to be spelled out a little more, essentially you have claimed that 5x = 0, and since 5 is not equal to 0 that x is equal to 0. There exist rings where you can multiply two nonzero numbers and get a zero, so you should probably explicitly prove that you're OK in Z7 unless this is something you've already done in your class
     
  4. Aug 22, 2013 #3

    Zondrina

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    Hint : You need to use the fact that :

    ##gcd(|\mathbb{Z}_5|, |\mathbb{Z}_7|) = 1##

    as well as Lagrange's theorem.
     
    Last edited: Aug 23, 2013
  5. Aug 24, 2013 #4
    Yes, I see what you mean. If I am in Z6, 3≠0 (mod 6) but from 3x=0 (mod 6) I can't conclude that x=0 (mod 6) as any x even would satisfy 3x=0 (mod 6). In the original problem I could say x=0 (mod 7) just because gcm(5,7)=1. I am not so sure how to prove what I've claimed, I'll check it. Thanks.
     
  6. Aug 25, 2013 #5

    mathwonk

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    prove: the only factorization of 5 as a product of a factor of 5 and a factor of 7 is 5.1.
     
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