Proof for: semidirect product of solvable groups is solvable Do you know the proof for the following statement or where I can find it?
Semidirect product of solvable groups is solvable?

I thought that this property is so general that it should somewhere on the Net, but unfortunately I didn't find the answer.
Thanks
 A: Let $K,H$ be solvable groups and suppose $G=K\rtimes H$.  There's a subnormal series $$1=K_0\unlhd K_1 \unlhd \cdots \unlhd K_n=K$$ such that $K_{j+1}/K_j$ is abelian for each $j$.  By construction, $K\unlhd G$ and $G/K\cong H$, so $G/K$ has a subnormal series  $$1_{G/K}=X_0\unlhd X_1 \unlhd \cdots \unlhd X_m=G/K$$
with $X_{j+1}/X_j$ is abelian for each $j$.  By the lattice theorem, each $X_j$ is the image of a subgroup $Y_j$ of $G$ containing $K$ under the canonical homomorphism (in other words, $X_i=Y_j/K$). In particular, this lifts to a subnormal series $$\begin{array}{rcccccl}
K=Y_0&\unlhd& Y_1& \unlhd& \cdots& \unlhd& Y_m=G\\
\uparrow & & \uparrow & & & & \uparrow \\
1_{G/K}=X_0&\unlhd& X_1& \unlhd& \cdots& \unlhd& X_m=G/K\end{array}$$
We can thus glue the subnormal series of $K$ together with this lifted subnormal series to obtain the following subnormal series of $G$: $$1=K_0\unlhd K_1 \unlhd \cdots \unlhd K_n=K =Y_0\unlhd Y_1 \unlhd \cdots \unlhd Y_m=G$$
We know that $K_{j+1}/K_j$ is abelian, so it remains to check if $Y_{j+1}/Y_j$ is abelian.  By the third isomorphism theorem, $$Y_{j+1}/Y_j\cong (Y_{j+1}/K)/(Y_{j}/K)=X_{j+1}/X_j$$ which we assumed was abelian.  This is a subnormal series of $G$ with abelian quotients, so $G$ is solvable.
