On the (finite) Boolean lattice there is a group structure given by the symmetric difference and this group is an elementary abelian 2-group.

Question: Does there exist a natural group structure on general (finite) distributive lattices?

Other examples of a group structures for a given lattice would also be interesting.

  • $\begingroup$ Hard to see how there could be something like symmetric difference for order ideals... $\endgroup$ Commented Jun 24, 2020 at 13:39
  • $\begingroup$ Would you mind clarifying what you mean by "natural"? If you mean that the commutative monoid given by the join is an (abelian) group, then the answer is that there are no examples other than the lattice with 1 element (the trivial group $\mathbb{Z}/1\mathbb{Z}$) and the lattice with 2 elements (the group $\mathbb{Z}/2\mathbb{Z}$). Indeed, for any other bounded (distributive) lattice you can find an element $z \neq 0$ such that $1 \vee z = 1 = 1\vee 0$, which means that $\vee$ cannot be cancellative. Not specifically what you asked, but a consideration to keep in mind. $\endgroup$ Commented Jun 24, 2020 at 13:50
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    $\begingroup$ @StefanoGogioso: well already the example of the Boolean lattice and symmetric difference is different than the join operation. $\endgroup$ Commented Jun 24, 2020 at 13:51
  • $\begingroup$ @SamHopkins Thank you: I know. I was just trying to clarify that cases to do with extending the lattice operations have some limitations. I edited my comment to make that clear. $\endgroup$ Commented Jun 24, 2020 at 13:53
  • $\begingroup$ It occurs to me that orthomodular lattices might be an interesting generalistion for you to look at, because people have studied various notions of symmetric difference there in the years. For example, a partial (negative) result for a certain definition of symmetric difference in OMLs is given by this paper. where the author shows that the symmetric differences they define can only be associative or cancellative when the orthomodular lattices are actually boolean algebras. $\endgroup$ Commented Jun 24, 2020 at 14:36

2 Answers 2



If it's natural, it should be invariant under the automorphism group of the original lattice.

let $X$ be the free distributive lattice on 2 generators $x,y$: it has 6 elements, $$0\quad<\quad x\wedge y \quad<\quad \stackrel{x}{_y}\quad<\quad x\vee y\quad<\quad 1 $$ with $x,y$ not comparable. It has an automorphism exchanging $x$ and $y$, and fixing the other elements.

But no group of order 6 has no automorphism with this property (if $G$ is a finite group and an automorphism fixes $>|G|/2$ elements, it's identity, just because the set of fixed points is a subgroup).

Initial answer: no for arbitrary finite lattices (the following example is not distributive).

Consider the lattice of subgroups of the Klein group $C_2^2$. It has cardinal 5 (0, the whole plane and 3 lines), and the automorphism group (of order 6) has two fixed points and a 3-element orbit.

Now a group structure on 5 elements is cyclic of order 5 and its automorphism group is cyclic of order 4, so a group structure on 5 elements cannot be preserved by the original automorphism group of order 6.


This question has already been answered, but I will add a slightly different answer.

First, the question and YCor's answer both seem to assume that by a natural group structure we mean that the group is a reduct of the original structure. This means that the group operations are given as words in the original structure, as symmetric difference is given by the Boolean word $x\oplus y = (x\wedge \neg y)\vee (\neg x\vee y)$. I also make this assumption.

Any compatible relation on a structure is also a compatible relation of any reduct. (A relation $R\subseteq A^n$ is compatible on an algebraic structure $A$ if $R$ is a subalgebra of $A^n$.)

YCor's answer uses automorphisms to answer the question. The graph of an automorphism of $A$ is a compatible binary relation on $A$, so any automorphism of a distributive lattice will also be an automorphism of any reduct. Then YCor exhibits a nonidentity automorphism of a distributive lattice that fixes more than half the points, which is something that can't happen in a group.

But to me the more obvious relation to consider is the order relation $\leq$. Every nontrivial distributive lattice has a nondiscrete compatible order, while no nontrivial group has a nondiscrete compatible order. (Even nontrivial ordered groups do not have nondiscrete partial orders that are compatible with all group operations. Reason: $a\leq b$ for compatible $\leq$ implies $a^{-1}\leq b^{-1}$, which implies $a\cdot \underline{a^{-1}}\cdot b\leq a\cdot \underline{b^{-1}}\cdot b$, which implies $b\leq a$.)

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    $\begingroup$ No, I didn't assume that the group structure should be a reduct of the lattice structure. I assumed that it should be compatible with isomorphisms. $\endgroup$
    – YCor
    Commented Jun 24, 2020 at 21:30
  • $\begingroup$ Actually what you're saying just applies to finite total orderings, right? you're saying that there is no "natural" way to make every finite total ordered set a group. But of course it depends on "natural". Using reducts is very restrictive, and quite clearly from the 2-element total order viewed as lattice, we can't make a group law as reduct. But still identifying every finite total order to an initial segment of $\omega$ (nonnegative integers) endows it with a cyclic group law in a natural (= choice-free) way. $\endgroup$
    – YCor
    Commented Jun 24, 2020 at 21:41
  • $\begingroup$ No, I am not talking about finite total orderings. Any compatible partial ordering on any group is discrete. (A partial ordering $\leq$ is compatible with the operations of $A$ if it is a subalgebra of $A^2$.) $\endgroup$ Commented Jun 24, 2020 at 21:45
  • $\begingroup$ Finite total orderings are a subclass of finite distributive lattices. What I'm saying is that under your restrictive interpretation of "natural", it's already true that there is no "natural" group law on every finite total ordering. $\endgroup$
    – YCor
    Commented Jun 24, 2020 at 21:59

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