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Math Help - Minimal normal subgroups.

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    Minimal normal subgroups.

    Hi:

    In a certain book the following definition is given: Let G be a group. A normal subgroup N \neq 1 of G is a minimal normal subgroup of G if 1 and N are the only normal subgroups of G that are contained in N.

    Then the following proposition is stated: Let N be a minimal normal subgroup of G. If \varphi is an epimorphism from G to a group H, then N^\varphi = 1 or N^\varphi is a minimal normal subgroup of H.

    The author gives this proof: Let A \neq 1 be a normal subgroup of H that is contained in N^\varphi. Then A^{\varphi^{-1}}\cap N is a normal subgroup of G, and A^{\varphi^{-1}}\cap N \neq 1 since A \neq 1. Hence A^{\varphi^{-1}}\cap N = N and N^\varphi = A.

    OK. But why A \neq 1 \Rightarrow A^{\varphi^{-1}}\cap N \neq 1? Since \varphi is homomorphism and A normal in H, A^{\varphi^{-1}} is normal in G. And the intersection of two normal subgroups is a normal subgroup, that is A^{\varphi^{-1}}\cap N is normal in G. But A^{\varphi^{-1}}\cap N  \subseteq N and N is minimal normal in G. Hence either A^{\varphi^{-1}}\cap N =  1 or A^{\varphi^{-1}}\cap N =  N. Hence, if it were true that A^{\varphi^{-1}}\cap N \neq 1, we would have A^{\varphi^{-1}}\cap N = N. Assuming the latter is true, we have the following: N \subseteq A^{\varphi^{-1}}, N^\varphi \subseteq (A^{\varphi^{-1}})^\varphi. But since \varphi is onto, (A^{\varphi^{-1}})^\varphi = A. Hence N^\varphi \subseteq A. But A \subseteq N^\varphi by assumption. Hence N^\varphi = A.

    As you see, I can follow the steps in the proof, except for the inference " A \neq 1 \Rightarrow A^{\varphi^{-1}}\cap N \neq 1". Any suggestion?

    The book is The Theory of Finite Groups, An Introduction, by Kurzweil and Stellmacher, and the proposition is proposition 1.7.1, p. 36.
    Last edited by ENRIQUESTEFANINI; March 19th 2012 at 06:17 AM.
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