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

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

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

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

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

    Case $\displaystyle G = G_1 \times G_2$: 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; Jan 16th 2012 at 08:26 PM.
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