Rank of finite solvable group I am very interested in the following question. 
Is there a finite solvable group G with the property that rank G - rank G_ab > n for n > 2? Here G_ab denotes the abelianization of G. For all the examples I know, the difference of ranks is either 0, 1 or 2. I am trying to see if there exists a sequence of finite solvable groups with the property that the difference of ranks described above gets unbounded.
Dear Derek Holt and YCor, thank you very much for your answers. I would like to ask you a question about them. How did you compute the minimal number of generators of G in your answers? 
 A: Let $H$ be elementary abelian of order $p^n$ for an odd prime $p$, and let $G = H \rtimes \langle t \rangle$, with $t^2=1$, where $t$ inverts every element of $H$. (This is sometimes called a generalized dihedral group.) Then generating sets for $G$ have size at least $n+1$, but $|G/[G,G]|=2$.
To prove the claim about minimal generating sets, suppose that $X$ is a generating set of $G$. If $X$ contains more than two generatos $x,y \in G \setminus H$, then we can replace $y$ by $yx$ to get $y \in H$, so we may assume that there is a unique such generator. But, since all elements of $G \setminus H$ are of order $2$ and conjugate to each other, we may assume that $t \in X$. Let $|X| = m+1 $, where $m = |X \cap H|$.
Now $H$ is elementary abelian, so $|\langle X \cap H \rangle| \le p^m$. Also, since $t$ inverts every element of $H$, it normalizes all subgroups of $H$, and hence $|\langle X \rangle| \le 2p^m$. So $m \ge n$ and $|X| \ge n+1$.
Conversely, the union of $t$ and a minimal generating set of $H$ is a generating set of $G$ of size $n+1$.
A: Consider $K^*\ltimes K^n$ with $K^*$ acting by scalar multiplication on $K$, where $K$ is  a finite field with at least 3 elements. Its rank (I assume you mean minimal number of generators and it would have been useful to write it, there are other definitions) is $n+1$; the rank of its abelianization is 1.
Proof of the statement on the rank. It's clearly at most $n+1$ (since $K^*$ is cyclic. It is also easy to see that it is at least $n$. Indeed, if we have elements $(t_i,v_i)_{1\le i\le k}$, then the subgroup they generate is contained in $K^*\ltimes V$, where $V$ is generated by the $v_i$; in particular if $k\le n-1$ then they do not generate.
Actually it is $n+1$. The reason is that any element not in $K^n$ has order coprime to $|K|$, and hence is conjugate to an element of $K^*$ (by using basic results on Hall subgroups in finite solvable groups, or by a simple direct verification). Hence with no restriction we can suppose $v_0=0$, so by the previous argument we get $k\ge n+1$.
