Invariants of a finite abelian group written as a central extension of a cyclic group by a finite abelian group.

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Notation : If $A$ is a finite abelian group then $(d_r,...,d_1)$ are the invariants of $A$ if $d_r>1$ :

$$A\text{ is isomorphic to } \mathbb{Z}/d_r\times...\times \mathbb{Z}/d_1 \text{ and } d_{i+1}\text{ divides } d_i\text{ for } 1\leq i\leq r-1$$

My problem : If $A$ is a finite abelian group whose invariants are known, $C$ is a cyclic group of order $d$ and $B$ is an abelian group which is an extension of $C$ by $A$, i.e. we have the following (central) exact sequence :

$$1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1 $$

I would like to have all possible invariants of $B$ in terms of the invariants of $A$ and the order of $C$ that can arise this way.

For instance, if $r$ is the number of invariants of $A$ then the number of invariants of $B$ is $\geq r$, but since $B$ is generated by $r+1$ elements (the $r$ generators of $A$ plus one element that is sent onto a generator of $C$) we also have that $B$ has a number of invariants $\leq r+1$. So the number of invariants for $B$ is either $r$ or $r+1$. Furthermore some examples I made (on abelian $2$-groups) seem to point out that among the invariants of $B$ we have at least $r-1$ invariants of $A$... But I am not sure of this.

This problem seems natural to me and I am sure that it has been studied but I cannot find a reference for this on my own. If you could provide a reference for this, it would be great.

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The most general statement we get is, if $d$ is the exponent of $B$ then :

$$B\text{ is isomorphic to a subgroup of } A\times \mathbb{Z}/d $$

Remark that the isomorphism is absolutely non-trivial.