Why if H⊴G and H is solvable and G/H is solvable, then G is also solvable?

Question

Plus, to the proof any p-group G is solvable,although Z(G) is normal to G and which is abelian, but G/Z(G) is not abelian, so what is the chain subgroup to show that G is solvable?

Answer 1

Hint:

$$H\;\;\text{solvable}\;\implies\;\exists\;\;\text{an abelian series}\;\;1:=H_0\lhd H_1\lhd\ldots\lhd H_k:=H$$

$$G/H\;\;\text{solvable}\;\implies\;\exists\;\;\text{an abelian series}\;\;\overline 1:=\overline G_0\lhd \overline G_1\lhd\ldots\lhd\overline G_m:=G/H$$

Use that $\;G_i=N_i/H\;$ , with $\;H\le N_i\le G\;$ , and $\;\overline G_i\lhd\overline G_{i+1}\iff N_i\lhd N_{i+1}$

Now check what happens with the series

$$1=H_0\lhd\ldots\lhd H\lhd N_1\lhd\ldots ...\lhd G$$