# Central factor is not quotient-transitive

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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 $H \le K \le G$ 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 $G$.

## 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.

## 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 $[G,K] \le K$, 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

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 $G$.