Is there a constant $K \in \mathbb{N}$ such that for every finite solvable group $G$, there exists a nilpotent subgroup $N \leq G$, and a subset $S \subseteq G$ with $|S| \leq K$, and $\langle N,S\rangle = G$ ?



Finite solvable groups are generated by a nilpotent subgroup + K elements?
No. Call the minimal cardinal of a subset $S$ such that $H\cup S$ generates $G$, the (generating) corank of $H$ in $G$. I claim that 



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> There exists solvable (actually metabelian) finite groups for which the smallest corank of a nilpotent subgroup is arbitrary large.
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Consider a nonempty finite index set $I$, of cardinal $n$. Consider for $i\in I$ copies $A\_i$ of the group with 2 elements, $B\_i$ copies of the group of 3 elements. Let $A\_i$ act nontrivially on $B\_i$ (but trivially on $B\_j$ for $j\neq i$). Let $m$ be another integer and consider $$G=G\_{m,n}=\left(\prod\_i A\_i\right)\ltimes \left(\prod\_i B\_i^m \right)=\prod\_i (A\_i\ltimes B\_i^m)\simeq (C\_2\ltimes\_{\pm} C\_3^m)^n.$$


Let $H$ be a subgroup of $G$. 


First assume that $H$ is a 2-group. Then conjugating and enlarging $H$, we can assume that $H$ equals the 2-Sylow $\prod A\_i$. Hence the corank of $H$ is equal to the minimal number of generators of $\prod B\_i^m$ as a $(\prod A\_i)$-module, which is bounded above (actually equal but we don't care) by the minimal number of generators of $C\_3^m$ as a $C\_2$-module, and this number is clearly $m$ (the $C\_2$-action in this case being scalar, it is useless to reduce the generating rank).


Let now $H$ be arbitrary.
Let $J\subset I$ be the set of $i$ such that the projection $H\to A\_i$ is onto, and $K$ its complement. Then looking at the abelianization shows that the corank of $H$ is $\ge \#(K)$. Now assume that $H$ is nilpotent. Denote by $N$ the 3-torsion in $G$, which is a subgroup (the unique 3-Sylow). Then $N\cap H$ is the 3-Sylow in $H$; it is contained in $\prod\_{i\in K}B\_i^m$ (otherwise its projection on $B\_j^m$ is nonzero for some $j\in J$ and having in mind that in $H$, elements of order 2 commute with elements of order 3 as in any nilpotent group, we contradict the fact that the projection of $H$ on $A\_j$ is nontrivial since the action of the nontrivial element of $A\_j$ on $B\_j^m$ fixes only zero). Thus after killing $\prod\_{i\in K}A\_i\ltimes B\_i^m$, we are in the case when $H$ is a 2-subgroup, and we see that the corank of $H$ is at least equal to $m$, provided that $J$ is not empty. If $J$ is empty, then $n=\#(K)$ and hence the corank of $H$ is at least $n$.


This proves that the corank of any nilpotent subgroup $H$ of $G\_{m,n}$ is at least $\min(m,n)$. 


