# 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.