Let $H$ be a Krull monoid with finite class group $G$ and suppose that each class contains a prime divisor (rings of integers in algebraic number fields share this property). For each element $a \in H$, its set of lengths $\mathsf L(a)$ consists of all $k \in \mathbb{N} _0$ such that $a$ can be written as a product of $k$ irreducible elements. Sets of lengths of $H$ are finite nonempty subsets of the positive integers, and we consider the system $\mathcal L (H) = \{ \mathsf L (a) \mid a \in H \}$ of all sets of lengths. It is classical that H is factorial if and only if $|G| = 1$, and that $|G| \le 2$ if and only if $|L| = 1$ for each $L \in \mathcal L(H)$ (Carlitz, 1960).

Suppose that $|G| \ge 3$. Then there is an $a \in H$ with $|\mathsf L (a)|>1$, the $m$-fold sumset $\mathsf L(a) + \ldots +\mathsf L(a)$ is contained in $\mathsf L(a^m)$, and hence $|\mathsf L(a^m)| > m$ for every $m \in \mathbb{N}$. The monoid $\mathcal B (G)$ of zero-sum sequences over $G$ is again a Krull monoid of the above type. It is easy to see that $\mathcal L (H) = \mathcal L \big(\mathcal B (G) \big)$, and it is usual to set $\mathcal L (G) := \mathcal L \big( \mathcal B (G) \big)$. In particular, the system of sets of lengths of $H$ depends only on $G$, and it can be studied with methods from additive combinatorics.
The present talk is devoted to the inverse problem whether or not the class group $G$ is determined by the system of sets of lengths. In more technical terms, let $G'$ be a finite abelian group with $|G'| \ge 4$ and $\mathcal L(G) = \mathcal L(G')$. Does it follow that $G$ and $G'$ are isomorphic ?
The answer is positive for groups $G$ having rank at most two $[1]$ and for groups of the form $G = C_{n}^{r}$ with $r \le (n+2)/6$ $[2]$. The proof is based on the characterization of minimal zero-sum sequences of maximal length over groups of rank two, and on the set $\triangle^*(G)$ of minimal distances of $G$ (the latter has been studied by Hamidoune, Plagne, Schmid, and others ; see the talk by Q. Zhong).[-]