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Math Help - Homomorphism proof

  1. #1
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    Homomorphism proof

    If  f:G \rightarrow H and  x \in G has order k, prove that  f(x) \in H has order m, where m|k.

    Also, if gcd(|G|,|H|)=1, then  f(x)=1 for all  x \in G
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  2. #2
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    Hi

    You know that f is a group homomorphism, and that x has order k. Let's use this:
    f(e_G)=f(x^m)=f(\underbrace{x...x}_{\text{m times}})=\underbrace{f(x)...f(x)}_{\text{m times}}=f(x)^m (where e_G is the identity element of G)

    Moreover, since f is a homomorphism, f(e_G)=e_F.

    So we get f(x)^m=e_F. What can you deduce?


    Now, the second question. In a group, elements orders always divide the group order. If \text{gcd}(|G|,|H|)=1, then the only common positive divisor between |G| and |H| will be 1. For any x in G, we want to prove that f(x)=e_F, i.e. that f(x) has order 1. Some idea? (Using the first result of course)
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  3. #3
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    Quote Originally Posted by didact273 View Post
    If  f:G \rightarrow H and  x \in G has order k, prove that  f(x) \in H has order m, where m|k.

    Also, if gcd(|G|,|H|)=1, then  f(x)=1 for all  x \in G
    Since m|k, k=mx where x is a positive integer.
    By Lagrange's theorem, mx | |G| and m | |H|. Since |G| and |H| are coprime, there is no common factor other than 1. Thus, m is 1. We conclude that  f(x)=1 for all  x \in G.
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