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 be finite. Suppose that every maximal subgroup of is simple and normal in . Then is an Abelian group and , where are primes.
I proved this: Either has exactly one maximal subgroup or is a direct product of two maximal subgroups. And went on:
Case only one max: let it be . Then cyclic, cyclic. And now I must use an elementary statement from another book, which bothers me most, and is this:
, cyclic. Then . (1)
Accordingly, and are the only subgroups in . Then . And by means of I discover that , and I'm done with this case.
Case : 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?
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.