Central factor is not quotient-transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., central factor) not satisfying a subgroup metaproperty (i.e., quotient-transitive subgroup property).
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Statement

It is possible to have groups $H\leq K\leq G$ 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 $G$.

Definitions used

Central factor

Further information: 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. Maximal among abelian normal implies self-centralizing in nilpotent

Proof

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 $[G,K]\leq K$, and since $G$ has class two, we have $[G,K]\leq [G,G]\leq Z(G)$. Thus, $[G,K]\leq K\cap Z(G)=H$. 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

Further information: dihedral group:D8

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

Further information: 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 $G$.