Let G is a non-abelian solvable group and N is minimal normal subgroup of G. I want to show that |cd(G/N)| < |cd(G)| (Inequality is strict) .
Is it true? How could I show that?
We know this is not true for abelian group since if G is abelian then |cd(G/N)| = |cd(G)| =1
Or in a simple way, can we say that for a solvable group G , exists a minimal normal subgroup N such that|cd(G/N)| < |cd(G)| .
A minimal counterexample would have the property that $|G/N|$ is at least $6$, since it cannot be abelian either. Thus $|G|\geq 12$. The most obvious non-abelian group of order $12$, the dihedral group, is a counterexample. Generally, $D_{2^n}$ is a counterexample for $n\geq 4$.
More generally than $D_{12}$, $G\times A$, where $G$ is any non-abelian group and $A$ is an abelian group will do, where $N$ is a subgroup of $A$.