Series-equivalent abelian-quotient central subgroups not implies automorphic

Statement

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 $B\cong G/H\cong G/K$.

Proof

Example

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.

References

• Math Overflow question and answer (question by Vipul Naik, answer by Jack Schmidt)