It is clear that finite solvable groups are closed under those operations, so at most the solvable groups can be produced this way. Not all solvable groups can be written as semidirect products of abelian groups (for example, the quaternion group), but the quaternion group is a quotient of a semidirect product, so I'm wondering if all groups can be obtained using these methods.
I'm also curious if we can produce all solvable groups with less operations (for example, remove taking subgroups from our list), and whether if we throw in all simple groups, we can obtain all groups. Finally, I wonder if a bound can be obtained on how many times we'd need to use certain operations here (for example, is every finite solvable group a quotient of a single big iterated semidirect product?).
Best Answer
In fact, every solvable group can be obtained by a sequence of semidirect product and subgroup operators applied by starting with abelian groups. The trick is to start at top, rather than at the bottom, to avoid the issue I mentioned in the comments.
This relies on the theorem of Kaloujnine and Krasner that every extension of $G$ by $K$ can be embedded into the regular wreath product $G\wr K$. (Note: in my nomenclature, "extension of $G$ by $K$" means a group with a normal subgroup isomorphic to $G$ and corresponding quotient isomorphic to $K$). The regular wreath product is constructed by taking a direct product of $|K|$ copies of $G$, and then letting $K$ act on this product by acting on the coordinates via the regular action. Then you take the semidirect product $G^{|K|}\rtimes K$.
So proceed by induction on the solvability length. A solvable group of length $1$ is abelian. And a solvable group of solvability length $n+1$ can be realized as an extension of an abelian group $A$ by a group of solvability length $n$ $K$ (e.g., take the $n$th derivated group as the normal subgroup). This extension can be embedded into the wreath product $A\wr K$. Now, this is a semidirect product of an abelian group, $A^{|K|}$, by $K$; and $K$ can be obtained, inductively, as a sequence of subgroup and semidirect product operators with abelian groups. This proves that so can $G$.
Explicitly, if $G$ is solvable, let $n$ be the largest integer such that the $n$th derived subgroup $G^{(n)}$ is nontrivial. Start with $G^{\rm ab}=G/G'$; then construct the extension of $G'/G''$ by $G/G'$ by taking the corresponding subgroup of the wreath product $(G'/G'')\wr (G/G')$. This can be done under your restrictions because the base, $(G'/G'')^{|G/G'|}$, is actually abelian. Continue this way until you get to the extension of $G^{(n-1)}/G^{(n)}$ by $G/G^{(n-1)}$, where again the base is abelian.
You can certainly do this to obtain all finite groups if you allow simple groups, since a finite direct product that occurs in a base can be realized as a finite sequence of (trivial) semidirect products. But the trick of just looking at the base as an abelian group won't work if we are considering simple groups when the extension has an infinite quotient.