Series-equivalent not implies automorphic in finite abelian group

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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

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 \cong K and G/H \cong 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 = \{ (pa,0,b,c) \} = pC_{p^3} \times 0 \times C_p \times C_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.