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.)