Central factor is not quotient-transitive

Statement
It is possible to have groups $$H \le K \le G$$ such that:


 * $$H$$ is a central factor of $$G$$. (In particular, $$H$$ is normal in $$G$$).
 * $$K/H$$ is a central factor of $$G/H$$.
 * $$K$$ is not a central factor of $$G$$.

Central factor
A subgroup $$H$$ of a group $$G$$ is termed a central factor if $$HC_G(H) = G$$ where $$C_G(H)$$ is the centralizer of $$H$$ in $$G$$.

Note that any central subgroup, i.e., any subgroup contained in the center, is a central factor.

Related facts

 * Centrality is not quotient-transitive
 * Direct factor is quotient-transitive
 * Central factor over direct factor implies central factor

Facts used

 * 1) uses::Maximal among abelian normal implies self-centralizing in nilpotent

A generic example
Let $$G$$ be a finite non-Abelian group of nilpotence class two and $$K$$ be maximal among Abelian normal subgroups of $$G$$. Consider $$H = K \cap Z(G)$$. Then:


 * $$H$$ is a central factor of $$G$$: In fact, $$H$$ is a central subgroup of $$G$$ -- it is contained in the center of $$G$$.
 * $$K/H$$ is a central factor of $$G/H$$: Since $$K$$ is normal in $$G$$, we have $$[G,K] \le K$$, and since $$G$$ has class two, we have $$[G,K] \le [G,G] \le Z(G)$$. Thus, $$[G,K] \le K \cap Z(G) = H$$. Thus, $$[G/H,K/H]$$ is trivial, so $$K/H$$ is central in $$G/H$$. Thus, $$K/H$$ is a central factor of $$G/H$$.
 * $$K$$ is not a central factor of $$G$$: By fact (1), $$K$$ is a self-centralizing subgroup of $$G$$. Since $$K$$ is proper, this yields that $$K$$ is not a central factor of $$G$$.

Example of the dihedral group
Consider the dihedral group:

$$G := \langle a,x \mid a^4 = x^2 = e, xax^{-1} = a^{-1} \rangle$$

Define:

$$H = \langle a^2 \rangle, K = \langle a \rangle$$.

Then, we have:


 * $$H$$ is a central factor of $$G$$: In fact, $$H$$ equals the center of $$G$$.
 * $$K/H$$ is a central factor of $$G/H$$: In fact, $$G/H$$ is Abelian, so any subgroup of it is a central factor.
 * $$K$$ is not a central factor of $$G$$: $$K$$ is a proper self-centralizing subgroup of $$G$$, so is not a central factor of $$G$$.

Example of the quaternion group
Consider the quaternion group, with identity element $$1$$, and with the list of elements:

$$G = \{ \pm 1, \pm i, \pm j, \pm k \}$$.

Define:

$$H = \langle -1 \rangle, K = \langle i \rangle$$.

Then, we have:


 * $$H$$ is a central factor of $$G$$: In fact, $$H$$ equals the center of $$G$$.
 * $$K/H$$ is a central factor of $$G/H$$: In fact, $$G/H$$ is Abelian, so any subgroup of it is a central factor.
 * $$K$$ is not a central factor of $$G$$: $$K$$ is a proper self-centralizing subgroup of $$G$$, so is not a central factor of $$G$$.