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Thread: G, H groups, phi: G -> H epimorphism, N normal in G, A normal in H. Then ...

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    G, H groups, phi: G -> H epimorphism, N normal in G, A normal in H. Then ...

    Hi: Let G be a group and N normal in G such that 1 and N are the only normal subgroups of G that are contained in N. Let $\displaystyle \phi$ be a surjective (onto) homomorphism from G to a group H. Let $\displaystyle A \neq 1$ be a normal subgroup of $\displaystyle H$ that is contained in $\displaystyle N^\phi$ and let $\displaystyle M = A^{\phi^{-1}} \cap N$. Then $\displaystyle M$ is normal in $\displaystyle G$.

    OK. But I am said that, then, $\displaystyle M \neq 1$ since $\displaystyle A \neq 1$. I have been trying hard to prove that $\displaystyle M \neq 1$ but I could not. Any suggestion?

    Notation: 1 stands for the trivial subgroup of G.
    Last edited by ENRIQUESTEFANINI; Mar 20th 2012 at 04:14 PM.
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    Re: G, H groups, phi: G -> H epimorphism, N normal in G, A normal in H. Then ...

    Hi enriquestefanini,

    well, by the definition of a function $\displaystyle f: G \rightarrow H $, and you have g $\displaystyle \in$ G,
    then there is at most one image f(g) = h $\displaystyle \in$ H (there cannot be two distinct elements of H being
    the image of g, since this would be in contrary to the definition of a function).

    So if you have at least two elements in the image of f, e.g. $\displaystyle im(f) = \{h_1 , h_2\} $, then
    you have at least two different elements of G. Then it follows that M cannot be 1 alone, since if it was,
    there would be only one element in f(M).

    Further, the image f(N) $\displaystyle \subset $ H.

    Is that clear to you now?
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    Re: G, H groups, phi: G -> H epimorphism, N normal in G, A normal in H. Then ...

    Quote Originally Posted by mastermind2007 View Post
    Hi enriquestefanini,

    well, by the definition of a function $\displaystyle f: G \rightarrow H $, and you have g $\displaystyle \in$ G,
    then there is at most one image f(g) = h $\displaystyle \in$ H (there cannot be two distinct elements of H being
    the image of g, since this would be in contrary to the definition of a function).

    So if you have at least two elements in the image of f, e.g. $\displaystyle im(f) = \{h_1 , h_2\} $, then
    you have at least two different elements of G. Then it follows that M cannot be 1 alone, since if it was,
    there would be only one element in f(M).

    Further, the image f(N) $\displaystyle \subset $ H.

    Is that clear to you now?
    i don't think this alone by itself is enough. surely the pre-image of A has more than one element in it, but we need to show more: that φ-1(A)∩N cannot be trivial.

    but it's really deceptively simple: since A is non-trivial, we have a ≠ 1 in A.

    since A is contained in φ(N), we have a = φ(n), for some n in N.

    this n is in φ-1(A)∩N, and is a non-identity element.
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    Re: G, H groups, phi: G -> H epimorphism, N normal in G, A normal in H. Then ...

    Thanks very much, guys. It's pretty clear now.
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