Series-equivalent not implies automorphic in finite abelian group

Statement

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 $K$ ).

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

We construct an example of an abelian group $G$ of order $p^{7}$ , and subgroups $H$ and $K$ of order $p^{4}$ such that $H ≅ K$ and $G / H ≅ G / K$ .

We denote by $Z_{n}$ the group of integers modulo $n$ .

$G : = Z_{p^{3}} \times Z_{p^{2}} \times Z_{p} \times Z_{p}$ .

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

$H = {(p a, 0, b, c)} = p Z_{p^{3}} \times 0 \times Z_{p} \times Z_{p}$ .

$K = {(p a, p b, a, c)}$ .

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, $H ≅ K$ and $G / H ≅ G / K$ .

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.