I understand that the only free groups that are abelian are 1 and Z, hence a difference between 'free abelian groups' and 'abelian free groups'.
Can someone please tell me what are the solvable free groups?
How can I construct 'free solvable groups'?
What is known about free groups in other classes of groups like polycyclic, nilpotent ...?
Thanks.
On
There is no "free solvable group", but there are "free solvable groups of length $n$" for any $n$. There are also free nilpotent groups of class $n$, free Burnside groups of exponent $k$; all groups that are abelian-by-exponent $k$; and more.
More precisely: a variety of groups is a class of groups that is closed under taking isomorphisms, subgroups, quotients, and arbitrary direct products (that is, if $G\in\mathcal{C}$ and $K\cong G$, then $K\in\mathcal{C}$; if $G\in\mathcal{C}$ and $H\lt G$, then $H\in \mathcal{C}$; if $G\in\mathcal{C}$ and $N\triangleleft G$, then $G/N\in\mathcal{C}$; and if $\{G_i\}$ is an arbitrary family of groups with $G_i\in\mathcal{C}$ for every $i$, then $\prod G_i\in\mathcal{C}$). Examples of varieties include "all abelian groups", "all groups of solvability length at most $n$"; "all nilpotent groups of class at most $c$"; "all groups such that every element is of exponent $k$"; and more.
If $\mathcal{C}$ is a variety of groups, and $G$ is any group, then there is a smallest normal subgroup $\mathcal{C}(G)$ of $G$ such that $G/\mathcal{C}(G)\in\mathcal{C}$ (this normal subgroup is, in fact, not merely normal, but fully invariant). For example, when $\mathcal{C}$ is the class of all abelian groups, $\mathcal{C}(G)$ is none other than the commutator subgroup of $G$. Proof of the existence of $\mathcal{C}(G)$: Let $\{N_i\}_{i\in I}$ be the collection of all normal subgroups of $G$ such that $G/N_i\in\mathcal{C}$; the class is nonempty, since $G/G$ is always in any variety. Then consider the obvious map from $G$ to $\prod G/N_i$; the product must be in $\mathcal{C}$ (it is a product of groups in $\mathcal{C}$), so the image of $G$ is in $\mathcal{C}$; the kernel of this map is the group $\mathcal{C}(G)$.
So if $X$ is any set, and $F(X)$ is the free group on $X$, then there is a smallest subgroup $\mathcal{C}(F(X))$ such that $F(X)/\mathcal{C}(F(X))\in\mathcal{C}$. The resulting quotient can be seen to have the same universal property relative to groups in $\mathcal{C}$ as $F(X)$ does relative to all groups. So we say that this quotient is the "relatively free $\mathcal{C}$-group on $X$." We call the original group $F(X)$ the "absolutely free group on $X$" to distinguish it.
Added: Just in case, remember that if $\mathcal{D}$ is a category in which the objects are sets and the arrows are maps of sets, then given a set $X$, the free $\mathcal{D}$-object on $X$ is an object $F(X)$ in $\mathcal{D}$, together with a set-theoretic inclusion $i\colon X\to F(X)$, such that for every object $D\in\mathcal{D}$ and every set-theoretic map $f\colon X\to D$ there exists a unique $\mathcal{D}$-morphism $\varphi\colon F(X)\to D$ such that $f=\varphi\circ i$. For the "absolutely free groups", the category is the category of all groups; for "free abelian groups", the category is the category of abelian groups. For "relatively free $\mathcal{C}$-group", the category is the category of all groups in $\mathcal{C}$.
The reason there is no "free solvable group" is that the class of all solvable groups is not a variety: it is not closed under arbitrary direct products. But if you put a bound on the solvability length then you do get a variety, and you can construct the corresponding "free solvable group of length $n$".
In addition you can take the pro-$\mathcal{C}$ approach that Matt E mentions.
If you are interested in learning more about varieties, these lie at the intersection of Universal Algebra and Group Theory. I would recommend Hanna Neumann's book Varieties of Groups (a bit old, but still the standard reference); also George Bergman's Universal Algebra notes (available here).
I think that you might have to bound the derived length to get precisely the notion that you want.
Notation: For any group $G$, let $G^{(i)}$ denote the $i$th iterated derived subgroup of $G$, so $G^{(0)} = G$ and $G^{(i+1)} = [G^{(i)},G^{(i)}]$). A group is solvable if $G^{(i)}$ is trivial for some $i$, and has derived length $d$ if $G^{(d)}$ is trivial, but $G^{(i)}$ is non-trivial for $i < d$.
Now let $F(X)$ be the free group on a non-empty set $X$ of generators, and consider $F(X)/F(X)^{(d)}$. On the one hand, this group is clearly solvable of derived length $d$. On the other hand, if $G$ is any solvable group of derived length $\leq d$, then giving a map of sets $X \to G$ will be equivalent to giving a map of groups $F(X)/F(X)^{(d)} \to G$. Thus $F(X)/F(X)^{(d)}$ is the free object on $X$ in the category of solvable groups of derived length $\leq d$.
One can make a similar construction if you consider e.g. the category of nilpotent groups of nilpotency class $\leq c$ (i.e. whose central series is of length $\leq c$).
If you really want to consider all solvable groups at once, then you can consider the projective limit $$\lim_{\leftarrow} F(X)/F(X)^{(d)},$$ the so-called pro-solvable completion of $F(X)$. Similarly, in the nilpotent setting, you can form the pro-nilpotent completion of $F(X)$. But now the "free object" is not literally in the category you want, but is instead is a projective limit of groups from that category. (This is often okay, though; is you search for "pro-nilpotent completion", you will see that pro-nilpotent completions of free groups come up a lot, e.g. in arithmetic geometry.)