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Let $N$ be a minimal normal subgroup of a solvable finite group $G$. We know that $N$ is an elementary abelian $p$-group. Is it true that $dl(G/N)=dl(G)-1$?

Where $dl$ means derived length.

If it's false I would like a counterexample, if it is true a proof.

EDIT: as Arturo said, it's false. There are some additional conditions that make it true? For example if we add the contidion $G$ not abelian?

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Let $G=C_2\times C_2$, $N=C_2\times\{1\}$. The derived length of $G/N$ is the same as the derived length of $G$, since they are both abelian. –  Arturo Magidin Mar 16 '12 at 4:07
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For a nonabelian example, just take a direct product of $C_2$ with any nonabelian group. Another example is the dihedral group of order 18, which has a unique minimal normal subgroup. You will have difficulty finding sensible conditions under which this is true! –  Derek Holt Mar 16 '12 at 8:41

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Dihedral 2-groups are a good example to keep in mind. They have a unique minimal normal subgroup (that has order 2 and coincides with the center), but unless they are abelian they have derived length exactly 2. For example, you can quotient out by the unique minimal normal subgroup 7 times without changing the derived length of a dihedral group of order 1024.

In other words, the smallest non-identity subgroup in the derived series can be arbitrarily large in terms of chief length.

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