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 $G / K$ ).

Related facts

Series-equivalent characteristic subgroups may be distinct

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 $C_{n}$ the cyclic group of order $n$ .

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

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

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

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

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.