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The finite solvable groups G whose commutator subgroup G' is nilpotent, look interesting to me (in view of the easy structure of finite nilpotent groups and the resemblance with solvable Lie algebras in char. 0). Let me call such groups perfectly solvable. Note that the finite supersolvable groups (groups with cyclic normal series) are perfectly solvable.

Have such groups been studied in Literature ? Thank you.

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    $\begingroup$ These are precisely the finite groups $G$ such that $G/F(G)$ is Abelian. Additionally, $G/F(G)$ is isomorphic to a subgroup of ${\rm Out}(F(G))$ for such a group. I'm not sure what more one can expect to say. $\endgroup$
    – Geoff Robinson
    Nov 2, 2016 at 12:35
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    $\begingroup$ I have no idea if these have actually been studied. One thing is that they satisfy the Taketa inequality, as I explain at math.stackexchange.com/questions/272849/… $\endgroup$
    – Tobias Kildetoft
    Nov 2, 2016 at 12:35
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    $\begingroup$ Another characterization is that they are the groups $G$ such that $G/\Phi(G)$ is Metabelian. $\endgroup$
    – Geoff Robinson
    Nov 2, 2016 at 13:16
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    $\begingroup$ Groups with a nilpotent derived subgroup are called "nilpotent-by-abelian". @GeoffRobinson I'm not sure everybody knows your notation ($F(G)$, $\Phi(G)$) by heart. I don't. $\endgroup$
    – YCor
    Nov 3, 2016 at 4:46
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    $\begingroup$ @YCor: $F(G)$ is the Fitting subgroup (unique largest normal nilpotent subgroup); $\Phi(G)$ is the Frattini subgroup (intersection of maximal subgroups). $\endgroup$
    – Arturo Magidin
    Nov 3, 2016 at 5:02

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