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Math Help - group theory

  1. #1
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    group theory

    Let f,g :\mathbb{Z}/5\mathbb{Z} \to S_{5} be 2 non trivial group homomorphisms. Prove that there exists a \sigma \in S_{5} such that f(x)=\sigma g(x) \sigma^{-1} for every x \in \mathbb{Z}/5\mathbb{Z}
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  2. #2
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    Quote Originally Posted by Chandru1 View Post
    Let f,g :\mathbb{Z}/5\mathbb{Z} \to S_{5} be 2 non trivial group homomorphisms. Prove that there exists a \sigma \in S_{5} such that f(x)=\sigma g(x) \sigma^{-1} for every x \in \mathbb{Z}/5\mathbb{Z}

    Hints:

    1) if f,g are non-trivial homomorphisms then automatically they're monomorphisms (i.e., \ker f\,,\,\ker g\neq {0}), so what must be their images?

    2) Lemma: two cycles in S_n are conjugate iff they both have the very same length.

    Tonio
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  3. #3
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    HI

    Hi--

    Tonio, is that the only way to go. Actually, i did this but i am not sure whether i am right or wrong.

    S=\{ f(x) | x \in \mathbb{Z}_{5}\}, \ R=\{\sigma g(x) \sigma^{-1}\ | x \in Z_{5}\}

    I actually gave a one-one map from S->R. Does that solve this by any means?

    --------

    Em, as for your question the yeah, their images must always be a 5-cycle in S_5
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Chandru1 View Post
    Hi--

    Tonio, is that the only way to go. Actually, i did this but i am not sure whether i am right or wrong.

    S=\{ f(x) | x \in \mathbb{Z}_{5}\}, \ R=\{\sigma g(x) \sigma^{-1}\ | x \in Z_{5}\}

    I actually gave a one-one map from S->R. Does that solve this by any means?
    No...I mean, what if you were mapping into an abelian group instead of S_5? What you proved would still hold, but the result you are trying to prove would not.
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  5. #5
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    dindnt get

    Could you be a bit more precise.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Chandru1 View Post
    Could you a bit more precise.
    With what you should do, or with the problem with your proof?
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  7. #7
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    HI

    with the problem in my proof
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Chandru1 View Post
    with the problem in my proof
    You haven't done anything, really. You have said that there is a bijection between the set of images and their conjugates under fixed \sigma (although if the \sigma was variable - as in the set R is actually R = \cup R_{\sigma} where R_{\sigma} = \{\sigma^{-1} f(x) \sigma : x \in \mathbb{Z}_5\} then you would be fine).

    However, what you have proven will always hold although the result you are trying to prove does not hold. You have proven A and you want to show B, but A \not\Rightarrow B as there exist counter-examples.
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  9. #9
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    Quote Originally Posted by Chandru1 View Post
    Let f,g :\mathbb{Z}/5\mathbb{Z} \to S_{5} be 2 non trivial group homomorphisms. Prove that there exists a \sigma \in S_{5} such that f(x)=\sigma g(x) \sigma^{-1} for every x \in \mathbb{Z}/5\mathbb{Z}
    As tonio hinted, f and g are embeddings, i.e., the images of f and g are cyclic subgroups of order 5 in S_5.

    Let A be the image of f such that A=<(a_1, a_2, a_3, a_4, a_5)> and let B be the image of g such that B=<(b_1, b_2, b_3, b_4, b_5)>. Since the generator of A and B have the same cycle type, they are in the same orbit under the action of conjugation. That means, there exists \sigma \in S_5 such that \sigma b \sigma^{-1}=a, where b and a are generators of B and A, respectively.

    Once we find a \sigma, the bijection is established between B and A. Think of a \sigma relabels the generator of B to the generator A. Then each element of B corresponds to A in a well-defined manner.
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