Series-equivalent not implies automorphic in finite abelian group

Statement

In terms of subgroups

There can exist a finite abelian group $G$ and subgroups $H$ and $K$ of $G$ such that $H$ and $K$ are series-equivalent subgroups (in other words, $H$ is isomorphic to $K$ and the quotient group $G/H$ is isomorphic to the quotient group $G/K$) but are not automorphic subgroups (i.e., there is no automorphism of $G$ sending $H$ to $K$).

The smallest example for $G$ has order $2^6$, and a similar generic example can be constructed for $p^6$ for any prime number $p$.

In terms of extensions

There can be a pair of finite abelian groups $A$ and $B$ and two extensions with normal subgroup $A$ and quotient group $B$ such that:

1. The total groups in both extensions are abelian, and are isomorphic groups.
2. The two extensions are not pseudo-congruent extensions, i.e., they cannot be realized as equivalent to each other using automorphisms of $A$ and $B$.

In terms of cohomology and automorphisms

There can be a pair of finite abelian groups $A$ and $B$ and two elements $\sigma ,\phi$ are elements in the second cohomology group for trivial group action $H^2(B,A)$ such that:

1. $\sigma$ and $\phi$ are both represented by symmetric 2-cocycles, hence correspond to abelian group extensions.
2. The total groups of the group extensions obtained using the elements $\sigma$ and $\phi$ are isomorphic as groups.
3. $\sigma$ and $\phi$ are not in the same orbit of $H^2(B,A)$ under the action of $\mathrm{A}\mathrm{u}\mathrm{t}(A)\times \mathrm{A}\mathrm{u}\mathrm{t}(B)$.

Equivalence of formulations

• Between extensions and subgroups formulations: The formulation in terms of extensions can be interpreted in terms of subgroups as follows: in the first extension $A$ is realized as $H$ and $B$ as $G/H$, and in the second extension, $A$ is realized as $K$ and $B$ as $G/K$. The absence of an automorphism sending $H$ to $K$ is equivalent to the absence of a pseudo-congruence of extensions.
• Between cohomology and extensions formulations: Direct from the interpretation of the second cohomology group in terms of group extensions.

Related facts

Weaker facts

• series-equivalent not implies automorphic

Here are some intermediate versions:

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 abelian-quotient abelian not implies automorphic	$H$ and $G/H$ are both abelian	16	nontrivial semidirect product of Z4 and Z4	direct product of Z4 and Z2	cyclic group:Z2
series-equivalent characteristic central subgroups may be distinct	$H$ and $K$ are both central subgroups of $G$	32	SmallGroup(32,28)	cyclic group:Z2	direct product of D8 and Z2
series-equivalent abelian-quotient central subgroups not implies automorphic	$H$ and $K$ are central and $G/H,G/K$ are abelian	64	semidirect product of Z8 and Z8 of M-type	direct product of Z4 and Z2	direct product of Z4 and Z2

The notion of Hall polynomials

Further information: Hall polynomial

Hall polynomials are polynomials that give a formula for the number of subgroups in an abelian group of prime power order having a particular isomorphism class with a particular isomorphism class for the quotient group.

Proof

Example of order $p^6$

We construct an example of an abelian group $G$ of order $p^6$, and subgroups $H$ and $K$ of order $p^3$ such that $H\cong K$ and $G/H\cong G/K$.

We denote by ${\mathbb{Z}}_n$ the group of integers modulo $n$.

$G:={\mathbb{Z}}_{p^3}\times {\mathbb{Z}}_{p^2}\times {\mathbb{Z}}_p$.

We define the subgroups $H$ and $K$ as follows.

$H=\{(pa,0,b)\}=p{\mathbb{Z}}_{p^3}\times 0\times {\mathbb{Z}}_p\times {\mathbb{Z}}_p=⟨(p,0,0),(0,0,1)⟩$

$K=\{(pa,pb,a)\}=⟨(p,0,1),(0,p,0)⟩$

Note that the letter $a$ used in the definition of $K$ should be considered as an integer rather than an integer mod $p$, because its use for the first coordinate requires considering it mod $p^2$.

Then, $H$ and $K$ are both of type $(p^2,p)$, and the quotients $G/H$ and $G/K$ are both of type $(p^2,p)$. Thus, $H\cong K$ and $G/H\cong G/K$.

However, there is no automorphism of $G$ sending $H$ to $K$. For this, note that $H$ contains elements that are $p$ times elements of order $p^3$, but $K$ does not contain any such element.