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I have just studied super-solvable groups. The group $G$ is said super-solvable if $G$ has normal series $0=G_0\trianglelefteq G_1\trianglelefteq \cdots G_{n-1}\trianglelefteq G_n=G$ such that $\frac{G_i}{G_{i-1}}$ is cyclic for $i \in \{1,2,\ldots,n\}$.

I can show $G$ contains a normal maximal subgroup. Since group $\frac{G}{G_{n-1}}$ is cyclic, it has maximal normal subgroup $\frac{M}{G_{n-1}}$. We can deduce that $M\trianglelefteq G$ and $M$ is maximal subgroup of $G$.

Now, let $G$ be super-solvable group and non-abelian(not necessary finite). Does exist two distinct maximal normal subgroup of $G$?

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Not necessarily - cyclic groups of prime power order do not. –  Derek Holt Jan 25 '13 at 21:12
@DerekHolt, Isn't that an answer? –  JSchlather Jan 25 '13 at 21:20
@DerekHolt I've forgotten non-abelian. –  Babak Miraftab Jan 25 '13 at 21:29
You edited the question, adding the requirement for $G$ to be nonabelian. Now $S_{3}$ is an example of a nonabelian supersoluble group with just one maximal normal subgroup. –  Andreas Caranti Jan 25 '13 at 21:30

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Generally, this is not true. For example symmetric group of order $G=S_3$.

The group $G$ is super-solvable and $G$ has exactly one normal subgroup.

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"...has exactly one proper non-trivial normal subgroup"... –  DonAntonio Jan 25 '13 at 21:43

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