# Splitting a group into two subsets closed under multiplication

Given a group $$G$$, find subsets $$A,B$$ such that $$G=A\sqcup B$$ and $$A$$ and $$B$$ are closed under multiplication: $$x,y\in A$$ (corr. $$x,y\in B$$) implies $$xy\in A$$ (corr. $$xy\in B$$).

For example, if $$G$$ is finite then all splitting are trivial: if $$1\in A$$ then $$A=G$$ and $$B=\emptyset$$.

If $$G=\mathbb{Z}^d$$ then any splitting is determined by a halfspace $$\Pi^+$$ in $$\mathbb R^d\supset\mathbb Z^d$$: $$A=\Pi^+\cap\mathbb Z^d$$. (More precisely, by a flag of halfspaces $$\Pi^+\supset \Pi_1^+\supset\dots$$, because we can distribute the points of the hyperplane $$\partial\Pi^+$$ between $$A$$ and $$B$$.)

What about the free group $$G=F_d$$? Are there splitting which are not induced by the projection $$F_d\to\mathbb Z^d$$?

• You describe your motivation in a comment below, but, for future reference, such motivation should ideally go in the question itself. As is, the imperative ("find subsets …") and lack of context make this appear like a homework problem, and it could have been closed. Jul 24 at 20:28

This is a way to rediscover a quite well-studied class of groups:

Proposition Let $$G$$ be a group. Equivalent statements: (a) $$G$$ admits a partition $$G=A\sqcup B$$ with $$1\in A$$, $$B$$ nonempty, and both $$A,B$$ subsemigroups. (b) $$G$$ admits a nontrivial order-preserving action on some totally ordered set [which can be chosen to be the real line, or $$\mathbf{Q}$$, if $$G$$ is countable] (c) $$G$$ has a nontrivial left-orderable quotient.

The equivalence between (b) and (c) is classical. If $$G$$ has a non-trivial orderable quotient $$p:G\to Q$$, there is (by definition) a submonoid $$K$$ of $$Q$$ such that $$K\cap K^{-1}=\{1_Q\}$$ and $$K\cup K^{-1}=Q$$; pulling back to $$G$$ yields the desired decomposition.

Conversely, suppose that $$G=A\cup B$$ as above (with $$1\in A$$). Define $$H=A\cap A^{-1}$$: this is a subgroup (call it "core" of the decomposition).

Observe that $$HB=BH=B$$: indeed if $$h\in H$$ and $$b\in B$$, suppose by contradiction $$a:=hb\in A$$: so $$b=ah^{-1}\in AA\subset A$$, contradiction, same contradiction if $$bh\in A$$.

On $$G/H$$ define a total order by $$gH\le g'H$$ if $$g^{-1}g'\in A$$. Note that this doesn't depend on the choices of $$g,g'$$ modulo $$H$$ on the right. This relation is $$G$$-invariant, and is easily seen to be a total order. It is not reduced to a singleton (since $$H=A$$ would force $$B$$ empty, since the complement of a subgroup can't be closed under multiplication, unless empty).

Edit: this argument shows something more precise, not only describing the class of groups, but describing these decompositions:

Proposition 2: Let $$G$$ be a group. (1) If $$G$$ acts on a totally ordered set $$X$$ in a order-preserving and $$x\in X$$, define $$A=\{g\in G:gx\le x$$ and $$B=\{g\in G:gx>x\}$$. Then $$G=A\sqcup B$$ is a decomposition with the given axioms (and $$1\in A$$). (2) Conversely, any such decomposition of $$G$$ occurs in this way (for some such action and $$x$$, where we can assume in addition the action to be transitive).

Note that $$B$$ is empty iff $$G$$ fixes $$x$$.

Note that this shows that if $$G$$ has such a decomposition, it has one for which the core is a normal subgroup (while in general the core need not be normal).

Also, since non-abelian free groups are themselves left-orderable, they admit such decomposition that are not of the prescribed form (namely have such a decomposition with trivial core).

• So, the problem belongs to the filed of ordering of groups. My goal is to describe such splittings for the fundamental groups of surfaces including the free groups. Now I see I should look for actions of those groups on totally ordered sets. Thank you very much!
– nim
Jul 24 at 18:50
• Very nice answer, but I’m confused by your last observation, “…non-abelian free groups […] admit such decomposition that are not of the prescribed form”, which seems to contradict Prop. 2(2). Presumably I’m misunderstanding what you mean by “of the prescribed form” here; could you clarify that? Jul 25 at 7:50
• @PeterLeFanuLumsdaine "of the prescribed form" refers to OP's question, namely preimages of (some) orderings of the abelianization.
– YCor
Jul 25 at 9:48