I am looking for any efforts that have been made to characterize the character kernels (equivalently, the subgroups yielding cyclic quotients) inside the lattice of subgroups of a finite abelian group. I am looking for references, but if it happens that you know a complete description that could be given in an answer, of course I would be delighted with that too.

**Some thoughts:**

The set of character kernels is upward closed in the lattice of subgroups, as every quotient of a cyclic group is cyclic.

By the same token, for cyclic groups, every subgroup is a character kernel.

If $A = B\times C$ then the characters on $A$ are in bijection with pairs of characters on $B,C$; if $B,C$ furthermore have relatively prime exponent, then a character nontrivial on $b\in B$ and $c\in C$ is nontrivial on $(b,c)\in A$. Thus in this case the kernel of a character on $A$ is the direct product of the kernels of its restrictions to $B$ and $C$. The subgroup lattice is also the direct product. So (by considering arbitrary finite abelian $A$ as the direct product of its sylow subgroups) the problem is reduced to finite abelian $p$-groups.

For elementary abelian $p$-groups, the character group is the vector space dual so the character kernels are the hyperplanes, equivalently the maximal subgroups.

It is known (e.g. Butler,

*Subgroup lattices and symmetric functions*, Mem. Amer. Math. Soc. 112, no. 539) that the lattice of subgroups of an abelian $p$-group of type $\lambda = (\lambda_1,\dots,\lambda_\ell)$ (with $\lambda_1\geq \dots \geq\lambda_\ell$) surjects onto the lattice $L = [0,\lambda_1]\times\dots\times[0,\lambda_\ell]$ (where $[0,\lambda_i]$ is the chain of length $\lambda_i$). If a subgroup $B\subset A$ is a character kernel, is the same true of every subgroup that maps to the same node in $L$? If so, a characterization could be given by identifying the relevant nodes in $L$; this seems easy.

**Motivation:** (You can ignore this part, but comments on it are welcome.) For a problem in representation theory I am interested in determining which finite abelian groups have a proper nontrivial subgroup such that every character kernel contains some image of that subgroup under an automorphism of the whole group. Elementary abelian groups do have this, in fact any proper subgroup will do. The automorphism group acts transitively on the rank rows of the (ranked) subgroup lattice. The character kernels are precisely the maximal subgroups. If you pick any proper subgroup $B$, and any maximal subgroup $M$, the sublattice of subgroups of $M$ hits $B$'s rank row somewhere, so $B$ has automorphism images inside $M$. At the other extreme, cyclic groups never have this, since they have characters with trivial kernel. I've also got by hand that $C_p\times C_{p^k},\;k> 1$ never has such a subgroup: there are character kernels of order $p$ so no subgroup bigger than order $p$ will do; and the order $p$ subgroups all occur as character kernels but they are in two orbits under the automorphism group (one is contained in all the cyclic subgroups of order $>p$ and the rest are inside none of these); thus any subgroup fails to have automorphism images contained in the kernels from the other orbit. I suspect that at least for $p$-groups, $C_{p^{k_1}}\times\dots\times C_{p^{k_\ell}}$ can't have such a subgroup unless all the $k_i$'s are equal, and this is what I'm trying to figure out.