Let $G$ be a finite group. Suppose that we can write $G= A \rtimes B$ and also $A = C \rtimes D$. Further suppose that C is normal in $G$ (not just in $A$). Then can we write $G = C \rtimes E$ where $E=G/C$? Of course, if $|C|$ and $[G:C]$ are relatively prime, then this follows from the Schur-Zassenhaus theorem.


We can of course write $G= A \rtimes B = (C \rtimes D) \rtimes B$ and we would like to have some kind of "associativity" of semi-direct product so that 
$G= C \rtimes (D \rtimes B)$. The question is whether the semi-direct product "$D \rtimes B$" actually makes sense (that $D$ and $B$ act on $C$ is clear).


I am mainly interested in whether this works in the case that we have the following extra hypotheses (i) $G$ is soluble, (ii) $C$ is a $p$-group, (iii) $D$ is cyclic of order prime to $p$, and (iv) $B$ is cyclic (the question is only interesting if $p$ divides $|B|$ due to (iii) and Schur-Zassenhaus). We could also strengthen (i) to (v) $C$ is an elementary abelian $p$-group. I don't know whether any of these extra hypotheses are helpful or not. 


One can also phrase the problem in terms of splitting of short exact sequences. One has a commutative diagram 
$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$


$$
\begin{array}{c}
& & 1 & & 1 & & \\
& & \da{} & & \da{} & & \\
& & C & \ra{=} & C\\
 & & \da{} & & \da{} & & \\
1 & \ra{} & A & \ra{} & G & \ra{} & B & \ra{} & 1 \\
 & & \da{} & & \da{\pi} & & \da{=} \\
1 & \ra{} & D & \ra{} & E & \ra{} & B & \ra{} & 1 \\
 & & \da{} & & \da{} & & \\
& & 1 & & 1 & & 
\end{array}
$$
in which all rows and columns are exact. We are assuming that all the short exact sequences split (including the last row), apart from the one containing $\pi$; we want to show that this one also splits. It is easy to see what the section $\varepsilon: E \rightarrow G$ should be in terms of the other sections, but unfortunately I can't seem to show that it is actually a homomorphism. 


I have the feeling that there is either an easy solution or any easy counterexample satisfying the extra hypotheses above, so I am missing something either way. (I did try some small examples, which seemed to work.)



Iterated semi-direct products
I believe that it must split if $C$ is abelian.


More precisely, I will prove that if $C$ is an abelian $p$-group and $D$ is a $p'$-group, then $C$ has a complement in $G$ and so $G \cong C \rtimes (D \rtimes B)$. We are assuming that $G$ is finite. Note that under these assumptions $C$ is a characteristic subgroup of $A$, and hence normal in $G$.


There is a general result that says that, if $Q$ is a $p'$-group acting on a finite abelian $p$-group $P$, then $P = C\_P(Q) \times [P,Q]$. I don't have a reference to hand, but I know it is in Gorenstein's book on Finite Groups, and I can find the precise reference later. Note that this does not hold in general for nonabelian $p$-groups.


So we have $C=C\_C(D) \times [C,D]$.


Now $A/[C,D]$ is the direct product $C/[C,D] \times D[C,D]/[C,D]$. Since $A/D[C,D] \cong C/[C,D]$ is abelian, we have $[A,A] \le D[C,D]$ and so $[C,D] = C \cap [A,A]$ is normal in $G$. 


Since the direct factors $D[C,D]/[C,D]$ and $C/[C,D]$ of $A/[C,D]$ have coprime orders, they are both characteristic in $A/[C,D]$ and hence normal in $G/[C,D]$. In particular, $D[C,D]/[C,D]$ is normalized by the subgroup $B[C,D]/[C,D]$, which is a complement of $A/[C,D]$ in $G/[C,D]$.


Now the subgroup $BD[C,D]$ of $G$ has contains $B$ and $D$ and intersects $C$ in $[C,D]$. So it satisfies the original hypotheses, but with $C$ replaced by $[C,D]$. So if we now replace $G$ by $BD[C,D]$, we have $C\_C(D)=1$.


Now let $N = N\_G(D)$. Since, by the Schur-Zassenhaus Theorem, all complements of $C$ in $A$ are conjugate (to $D$), the Frattini Argument shows that $G=AN$,
and hence, since $D \le N$, $G=CN$. Also $[C \cap N,D] \le C \cap D = 1$, so $C \cap N \le C\_C(D) = 1$, and hence $C \cap N = 1$. So $N$ is a complement of $C$ in $G$. (It is also a complement of the original $C$ that we replaced by $[C,D]$.) 


Note that this argument doesn't use the facts that $D$ and $B$ are cyclic, but it does use the assumption that $C$ and $D$ have coprime orders.


I do not know whether such an extension always splits when $C$ is allowed to be a nonabelian $p$-group.


