Suppose I have a finite group G that is solvable. So we know this means $G^{(n)}$=1 for some $n \geq 0$. We can then write
$1=H_0 \triangleright H_1 \triangleright...\triangleright H_s=G $ such that $G^{(n-i)}=H_i$ and $H_{i+1}/H_i$ is abelian.
An equivalent definition for solvability would be:
$1=K_0 \triangleright K_1 \triangleright...\triangleright K_s=G$ and $K_{i+1}/K_i$ is cyclic.
Knowing $G^{(n-i)}=H_i$, is there a way to define $K_i$ in terms of $H_i$?
Well, you can always refine the first sequence. Assume that $G$ is finite and $H_{i+1}/H_i$ is abelian. Then by the Fundamental Theorem of Finite Abelian Groups
$$H_{i+1}/H_i \simeq C_{\alpha_1}\oplus\cdots\oplus C_{\alpha_k}$$
where each $C_{\alpha_j}$ is cyclic. By the third isomorphism theorem
$$C_{\alpha_j}\simeq H_{i+1}^j/H_i$$
for some subgroups $H_{i} < H_{i+1}^j < H_{i+1}$. Thus you can easily refine the sequence by putting
$$H_i < H_{i+1}^1 < H_{i+1}^1\oplus H_{i+1}^2 < \cdots < H_{i+1}^1\oplus\cdots\oplus H_{i+1}^k = H_{i+1}$$
Consecutive quotients are isomorphic to $C_{\alpha_j}$ thus they are cyclic.