# Character kernels in the lattice of subgroups of a finite abelian group

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.

• The answer to the question in the 5-th bullet point is "no", at least for the two surjections I can think of. And concerning your motivation, my guess is that an abelian $p$-group has no such subgroup iff all cyclic factors have different order. – Frieder Ladisch Mar 4 '16 at 19:48

A finite group is cyclic if and only if its subgroup lattice is distributive. (A more general result was proved by Baer.) So, if you are given the subgroup lattice of a finite group that you know to be abelian, you can find those subgroups yielding cyclic quotient using only the combinatorial structure of intervals in the lattice. (You cannot determine whether or not a group is abelian using only the combinatorial structure of its subgroup lattice.)

Roland Schmidt's "Subgroup Lattices of Finite Groups" is an excellent reference for questions about subgroup lattices.

• +1. If I am understanding you right, you are saying that if I have the subgroup lattice of a finite abelian group $A$, then the character kernels are exactly those nodes $x$ such that the upper interval $[x,A]$ is a distributive lattice. This is beautiful. – benblumsmith Feb 9 '17 at 16:37
• Although the other answer turned out to be exactly what I needed, this answer actually elegantly addresses the original question, so I feel it must be the one I accept. – benblumsmith Nov 10 '18 at 15:38

I doubt that there is any useful characterization of character kernels, other than that they have cyclic factor groups, but here is an answer to your motivating problem: Let $$A = C_{p^{k_1}} \times \dotsm \times C_{p^{k_d}}.$$ Then $A$ contains a subgroup $H$ such that every character kernel contains an $\DeclareMathOperator{\Aut}{Aut} \Aut(A)$-conjugate of $H$ if and only if there is $i\neq j$ with $k_i = k_j$.

If $A = C_{p^k} \times C_{p^k} \times B$ for some $B$, we can in fact take any $a \in C_{p^k}\times C_{p^k}$, $a\neq 1$, and set $H = \langle a \rangle$.

Now assume that $k_1 > k_2 > \dots > k_d$. For each $i$, let $z_i$ be an element of order $p$ in the $i$-th cyclic factor. For each $i$, the subgroup $Z_i:= \langle z_1, \dotsc, z_i \rangle$ is invariant under $\Aut(A)$. Suppose $H$ is a subgroup as required. Since the property is inherited by subgroups of $H$, we may assume that $H$ has order $p$ and thus $H \subseteq Z_d$. There is a character $\lambda$ such that $\ker(\lambda)\cap Z_d = Z_{d-1}$, and thus an $\Aut(A)$-conjugate of $H$ is contained in $Z_{d-1}$. Thus $H\leq Z_{d-1}$. There is also a character $\lambda$ such that $\ker(\lambda)\cap Z_d = \langle z_1, \dotsc, z_{d-2}, z_d\rangle$. It follows that an $\Aut(A)$-conjugate of $H$ is contained in $$\langle z_1, \dots, z_{d-2},z_d \rangle \cap Z_{d-1} = Z_{d-2}.$$ But this means that $H\leq Z_{d-2}$. Continuing in this way, we see that $H \leq Z_i$ for all $i$, and thus $H\leq Z_1 = \langle z_1 \rangle$. This is a contradiction since there is also a linear character not containing $z_1$ in its kernel.

• +1 and I'm marking this as accepted although it doesn't directly address the subgroup lattice question. – benblumsmith Mar 8 '16 at 19:16
• I am citing this argument in something. Did you compose it in response to this question? Or can you point me to a print reference? – benblumsmith Feb 8 '17 at 16:19
• @benblumsmith: If I remember correctly, I found this argument in response to your question. I don't know a reference for this. – Frieder Ladisch Feb 9 '17 at 12:18