Series-equivalent abelian-quotient abelian not implies automorphic

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:

$H$ 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.
$H$ 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 $K$ ).
$H$ 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

Series-equivalent not implies automorphic

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:

$G : = ⟨ x, y ∣ x^{4} = y^{4} = e, y x y^{- 1} = x^{3} ⟩$

Define $H$ as the subgroup $⟨ x^{2}, y ⟩$ and $K$ as the subgroup $⟨ x, y^{2} ⟩$ .

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.