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.

Related facts

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.