Series-equivalent not implies automorphic in finite abelian group

In terms of subgroups
There can exist a fact about::finite abelian group $$G$$ and subgroups $$H$$ and $$K$$ of $$G$$ such that $$H$$ and $$K$$ are fact about::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 fact about::automorphic subgroups (i.e., there is no automorphism of $$G$$ sending $$H$$ to $$K$$).

The smallest example for $$G$$ has order $$2^7$$, and a similar generic example can be constructed for $$p^7$$ 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.

Weaker facts

 * series-equivalent not implies automorphic

Here are some intermediate versions:

The notion of Hall polynomials
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.

Example of order $$p^7$$
We construct an example of an abelian group $$G$$ of order $$p^7$$, and subgroups $$H$$ and $$K$$ of order $$p^4$$ such that $$H \cong K$$ and $$G/H \cong G/K$$.

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

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

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

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

$$\! K = \{ (pa,pb,a,c) \}$$.

Then, $$H$$ and $$K$$ are both of type $$(p^2,p,p)$$, 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 $$(p^2,p,p)$$.