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Math Help - Abstract Algebra Proof Homomorphism

  1. #1
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    Abstract Algebra Proof Homomorphism

    Let G and H be two groups. If f: G \rightarrow H is a homomorphism, a \in G and b = f(a). If ord(a) = n, ord(b) = m, then n is a multiple of m. (Let e_{1} be the identity of G and e_{2} be the identity of H)

    I know a^n = ?
    and that b^n = ?

    I have to prove from that given that n is a multiple of m.

    Can I have some help with this proof? Thanks.
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  2. #2
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    Quote Originally Posted by Iceflash234 View Post
    Let G and H be two groups. If f: G \rightarrow H is a homomorphism, a \in G and b = f(a). If ord(a) = n, ord(b) = m, then n is a multiple of m. (Let e_{1} be the identity of G and e_{2} be the identity of H)

    I know a^n = ?
    and that b^n = ?
    Well, by definition of "order", you know that a^n= e_1 and b^n= e_2 (and there are no lower powers that give the identity).

    I have to prove from that given that n is a multiple of m.

    Can I have some help with this proof? Thanks.
    Apply f to both sides of a^n= e_1. Then f(a^n)= (f(a))^n= b^n= f(e_1)= e_2. Now you know that b^n= e_2 and that m is the smallest integer such that b^m= e_2.
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  3. #3
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    You lost me at the (f(a))^n = b^n. I know that f(a^n) = (f(a))^n but I don't see how that equals b^n and then how that equals f(e_1).
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  4. #4
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    Quote Originally Posted by Iceflash234 View Post
    You lost me at the (f(a))^n = b^n. I know that f(a^n) = (f(a))^n but I don't see how that equals b^n and then how that equals f(e_1).
    f(a) = b, so obviously, (f(a))^n = b^n. Now, a^n = e_1 \Rightarrow f(a^n) = f(e_1) = e_2
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  5. #5
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    I understand that now. Thank you.
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