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Math Help - If every maximal subgr of G is simple and normal then |G| = 1, p, p^2 or pq.

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    If every maximal subgr of G is simple and normal then |G| = 1, p, p^2 or pq.

    Hi:
    In the book The Theory of Finite Groups: An Introduction, by Hans Kurzweil and Bernd Stellmacher, New York, 2004, in section 1.6, Products of Groups, there is Exercise 3:

    Let G be finite. Suppose that every maximal subgroup of G is simple and normal in G. Then G is an Abelian group and  |G| \in \{1, p, p^2, pq\} , where p, q are primes.

    I proved this: Either G has exactly one maximal subgroup or G is a direct product of two maximal subgroups. And went on:

    Case only one max: let it be H. Then G cyclic, H cyclic. And now I must use an elementary statement from another book, which bothers me most, and is this:

    T \unlhd G, T cyclic. Then S \leq T  \Rightarrow S \unlhd G. (1)

    Accordingly, <1> and H are the only subgroups in H. Then |H| = p . And by means of G/H I discover that |G/H| = q, and I'm done with this case.

    Case G = G_1 \times G_2: for this case I tried to use commutators but I did not get anything. I think I could adapt the proof of (1), substituting the two conditions T maximal and T simple for T cyclic. But in proving (1) I directly manipulated elements of groups instead of groups, making an adaptation difficult.

    Could you give me some hint?

    EDIT:
    I can now prove (1) using only "global properties": Every subgroup of a cyclic group G is a characteristic subr of G. Hence, if S subgr of T, S is characteristic in T. But T normal in G. Therefor, S normal in G.
    Last edited by ENRIQUESTEFANINI; January 16th 2012 at 09:26 PM.
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