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 $\sigma,\varphi$ are elements in the second cohomology group for trivial group action $H^2(B,A)$ such that:

1. $\sigma$ and $\varphi$ 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 $\varphi$ are isomorphic as groups.
3. $\sigma$ and $\varphi$ are not in the same orbit of $H^2(B,A)$ under the action of $\operatorname{Aut}(A) \times \operatorname{Aut}(B)$.

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.

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

$\! K = \{ (pa,pb,a) \} = langle (p,0,1), (0,p,0) \rangle$

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.

Note on dual example

Since subgroup lattice and quotient lattice of finite abelian group are isomorphic, we can invert the above example so as to get both $H$ and $K$ of type $(p^2,p)$ and both $G/H$ and $G/K$ of type $(p^2,p,p)$.