Series-equivalent abelian-quotient abelian not implies automorphic

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Statement

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. H and G/H (and hence also K and G/K) are both abelian groups. In other words, H is an Abelian normal subgroup (?) as well as an Abelian-quotient subgroup (?) of G (and so is K).
  3. H and K are not Automorphic subgroups (?) in G, i.e., there is no automorphism of G sending H to K.

Related facts

Stronger facts

Statement Constraint on G,H,K Smallest order of G among known examples Isomorphism class of G Isomorphism class of H,K Isomorphism class of quotient group G/H, G/K
series-equivalent not implies automorphic in finite abelian group G is also a finite abelian group 128 direct product of Z8 and Z4 and V4 direct product of Z8 and V4 direct product of Z4 and Z2
series-equivalent abelian-quotient central subgroups may be distinct H and K are both central subgroups of G (in other words, the quotient actions are both trivial. 64 semidirect product of Z8 and Z8 of M-type direct product of Z4 and Z2 direct product of Z4 and Z2

Weaker facts

Proof

Example of the nontrivial semidirect product of cyclic groups of order four

Further information: nontrivial semidirect product of Z4 and Z4, subgroup structure of nontrivial semidirect product of Z4 and Z4

We define G as the nontrivial semidirect product of Z4 and Z4, explicitly by the presentation:

\! G := \langle x,y\mid x^4 = y^4 = e, yxy^{-1} = x^3 \rangle

Define H as the subgroup \langle x^2, y\rangle and K as the subgroup \langle x,y^2 \rangle.

Both H and K are isomorphic to direct product of Z4 and Z2, and they are both normal with the quotient group isomorphic to cyclic group:Z2 (because they both have index two).

To see that there is no automorphism of G sending H to K, note that the only non-identity squares of elements in the two subgroups are y^2 (in H) and x^2 (in K) respectively. An automorphism of G sending H to K must therefore send y^2 to x^2. This is not possible since x^2 is in the derived subgroup of G (it is the only non-identity commutator) and y^2 is not.