Series-equivalent abelian-quotient central subgroups not implies automorphic

From Groupprops
Jump to: navigation, search


In terms of series-equivalent subgroups

It is possible to have a finite group G (in fact, even a group of prime power order) and normal subgroups H and K of G such that:

  1. H and K are Series-equivalent subgroups (?) of G: H and K are isomorphic groups and the quotient groups G/H and G/K are also isomorphic groups.
  2. G/H (and hence also G/K) is an abelian group. In other words, H and K are both Abelian-quotient subgroup (?)s, i.e., they both contain the derived subgroup.
  3. H and K are both Central subgroup (?)s, i.e., they are both contained in the center of G.
  4. H and K are not Automorphic subgroups (?) in G, i.e., there is no automorphism of G sending H to K.

In terms of the second cohomology group

It is possible to have abelian groups A and B and elements of the Second cohomology group for trivial group action (?) H^2(B,A) that are not in the same orbit under the natural action of \operatorname{Aut}(B) \times \operatorname{Aut}(A) but nonetheless give isomorphic big groups.

Equivalence of statements

The statement for the second cohomology group can be interpreted in terms of series-equivalent subgroups by identifying A \cong H \cong K and B \cong G/H \cong G/K.



Further information: semidirect product of Z8 and Z8 of M-type

Let G be the semidirect product of Z8 and Z8 of M-type (GAP ID: (64,3)) given explicitly by the presentation:

\! G := \langle a,x \mid a^8 = x^8 = e, xax^{-1} = a^5 \rangle

The center of G is \langle a^2, x^2 \rangle and it is isomorphic to the direct product of Z4 and Z4. The derived subgroup of G is \langle a^4 \rangle, so any subgroup between these is an abelian-quotient central subgroup.

Let H be the subgroup \langle a^2, x^4 \rangle and K be the subgroup \langle a^4, x^2 \rangle. Both are central subgroups of G isomorphic to direct product of Z4 and Z2. In both cases, the quotient group is also isomorphic to direct product of Z4 and Z2. However, there is no automorphism sending H to K because K contains an element whose square is the unique non-identity commutator of G, while H does not.