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Math Help - Prove homomorphic image

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    Prove homomorphic image

    Suppose that G, G', and G'' are groups. If G' is a homomorphic image of G, and G'' is a homomorphic image of G', prove that G'' is a homomorphic image of G.

    Something tells me that it involve transitive property ...
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainyice View Post
    Suppose that G, G', and G'' are groups. If G' is a homomorphic image of G, and G'' is a homomorphic image of G', prove that G'' is a homomorphic image of G.

    Something tells me that it involve transitive property ...
    Obviously.

    So, we have \theta:G\to G',\theta':G'\to G'' which are both epimorphisms (surjective homomorphisms). So the question is why is \theta'\circ\theta:G\to G'' an epimorphism? But this is easy when put this way since the product of surjective maps is surjective and the product of homomoprhisms is a homomorphism. Prove those facts if you don't know them.
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    Quote Originally Posted by Drexel28 View Post
    Obviously.

    So, we have \theta:G\to G',\theta':G'\to G'' which are both epimorphisms (surjective homomorphisms). So the question is why is \theta'\circ\theta:G\to G'' an epimorphism? But this is easy when put this way since the product of surjective maps is surjective and the product of homomoprhisms is a homomorphism. Prove those facts if you don't know them.
    I get what you are saying, and I know how to prove it in term of function. When my professor did one problem of showing a group is one to one or onto ... I have difficulty understanding his strategy of proving it. Group theory is a brand new concept to me.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainyice View Post
    I get what you are saying, and I know how to prove it in term of function. When my professor did one problem of showing a group is one to one or onto ... I have difficulty understanding his strategy of proving it. Group theory is a brand new concept to me.
    Is there a question buried in there?
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    Quote Originally Posted by Drexel28 View Post
    Is there a question buried in there?
    Maybe you could tell me how to construct the prove by steps. It is obvious that G --> G'' is onto for all of their (G,G',G'') property or function is onto. However, how can I write it mathematically?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainyice View Post
    Maybe you could tell me how to construct the prove by steps. It is obvious that G --> G'' is onto for all of their (G,G',G'') property or function is onto. However, how can I write it mathematically?
    I mean you need to prove that the composition of surjective maps is surjective but this follows since if G\overset{\theta}{\longrightarrow}G'\overset{\thet  a'}{\longrightarrow}G'' each being a surjection then (\theta'\circ\theta)(G)=\theta'(\theta(G))=\theta'  (G')=G'' and so surjectivity follows.

    The homomorphism property follows since \theta'(\theta(gg'))=\theta'(\theta(g)\theta(g'))=  \theta'(\theta(g))\theta'(\theta(g'))
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    Quote Originally Posted by Drexel28 View Post
    Obviously.

    So, we have \theta:G\to G',\theta':G'\to G'' which are both epimorphisms (surjective homomorphisms). So the question is why is \theta'\circ\theta:G\to G'' an epimorphism? But this is easy when put this way since the product of surjective maps is surjective and the product of homomoprhisms is a homomorphism. Prove those facts if you don't know them.
    Another question, how do you know that the homomorphism from G to G' and G' to G'' are epimorphism?
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainyice View Post
    Another question, how do you know that the homomorphism from G to G' and G' to G'' are epimorphism?
    Because we know that G' is a homomorphic image of G
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