Locally finite groups and configurations

Let 𝐺 be a group. Two subsets 𝐴 and 𝐵 of 𝐺 are called equidecomposable if there exist partitions { A 1 , … , A n } and { B 1 , … , B n } of 𝐴 and 𝐵 respectively and elements g 1 , … , g n of 𝐺 such that

In such a case, we use the notation A ∼ B .

The equations Con ⁡ ( G 1 H 1 ) = Con ⁡ ( G 2 H 2 ) and Con ⁡ ( H 1 ) = Con ⁡ ( H 2 ) do not imply the equality Con ⁡ ( G 1 ) = Con ⁡ ( G 2 ) . For a counterexample, let G 1 = Z , H 1 = 2 ⁢ Z , G 2 = Z 2 × Z and H 2 = Z , see [2].

Let 𝐺 be a group, and let H ≤ G be a subgroup. Then, by [17, Theorem 3.8], the following holds.

1. Con ⁡ ( H ) ⊆ Con ⁡ ( G ) . In particular, τ ⁢ ( G ) ≤ τ ⁢ ( H ) .

2. If the subgroup 𝐻 is normal in 𝐺, then Con ⁡ ( G / H ) ⊂ Con ⁡ ( G ) . In particular, τ ⁢ ( G ) ≤ τ ⁢ ( G / H ) .

Let A , B , C , D be subsets of 𝐺 such that A ∩ B = ∅ . If A ∼ C and B ∼ D , then there exist two disjoint subsets 𝑀 and 𝑁 of A ∪ B such that ( C ∪ D ) ∼ M and ( C ∩ D ) ∼ N .

Proof

Let A = ⊔ i = 1 n A i , B = ⊔ j = 1 m B j , C = ⊔ i = 1 n C i , D = ⊔ j = 1 m D j , and for some g i ∈ G and h j ∈ G , we have

Then ( C ∪ D ) ∼ M and ( C ∩ D ) ∼ N , where

Let ( g , E ) be a configuration pair for 𝐺, where g = ( g 1 , … , g n ) is a generating string and E = { E 1 , … , E m } . Suppose 1 ≤ i 1 < i 2 < ⋯ < i s ≤ m , 0 ≤ j 1 , … , j s ≤ n and 0 ≤ k 1 , … , k s ≤ n are such that, for every C ∈ Con ⁡ ( g , E ) ,

and, for some C * ∈ Con ⁡ ( g , E ) ,

Then there exists B ⊆ G such that

Proof

As mentioned before, for every 1 ≤ i ≤ m and 0 ≤ j , k ≤ n ,

In particular,

Setting M j C = g j ⁢ x 0 ⁢ ( C ) , where x j ⁢ ( C ) ⊆ E i , we have

Consequently, for every 1 ≤ i ≤ m and 0 ≤ j ≤ n ,

Let D = ⋃ t = 1 s { C : x k t ⁢ ( C ) ⊆ E i t } . For C ∈ D , define

Since E i 1 , … , E i s are pairwise disjoint, by Lemma 2.3, there exist disjoint families of pairwise disjoint subsets { M 1 C , M 2 C , … , M p C C } C ∈ D of ⋃ t = 1 s E i t such that, for each C ∈ D ,

Now, by (2.2) and (2.3), there exist disjoint subsets M 1 , M 2 , … , M s , M * of ⋃ t = 1 s E i t such that x 0 ⁢ ( C * ) ∼ M * and

On the other hand, by (2.4), for every 1 ≤ t ≤ s ,

Therefore, ⊔ t = 1 s M t ⊊ ⊔ t = 1 s E i t , and however,

Is there an infinite group 𝐺 such that, for any Con ⁢ ( g , E ) in Con ⁡ ( G ) , there exists a finite group 𝐹 with Con ⁢ ( g , E ) ∈ Con ⁡ ( F ) ⁢ ?

Suppose 𝐺 is a finitely generated group, and let g = ( g 1 , … , g n ) be a generating string. The Cayley graph Γ = Cay ⁡ ( G , g ) is the directed graph with vertex set 𝐺 and edges the pairs ( x , g i ⁢ x ) ∈ G × G , where x ∈ G , i = 1 , 2 , … , n .

Let 𝐺 be finitely generated by the ordered set g = { g 1 , … , g n } . Then 𝐺 is infinite if and only if it possesses two subsets 𝐴 and 𝐵 such that A ⊊ B , which are equidecomposable.

Proof

The “if” part of the lemma is clear. Conversely, assume that 𝐺 is infinite. By the König lemma [4], the Cayley graph Cay ⁡ ( G , g ) contains a ray (a path with no repeated vertices) that starts at a vertex and continues from it through infinitely many vertices. It means that there exists an infinite subset B = { x 0 , x 1 , x 2 , … } of 𝐺 such that x j + 1 ⁢ x j - 1 ∈ g , j ∈ N ∪ { 0 } . Set A := B ∖ { x 0 } , and for 1 ≤ k ≤ n , define

Clearly, A ⊊ B . On the other hand, A = ⊔ k = 1 n A k and B = ⊔ k = 1 n B k . Therefore, 𝐴 and 𝐵 are equidecomposable. This proves the lemma. ∎

Let 𝐺 be a discrete group. Then 𝐺 is not locally finite if and only if it possesses two subsets 𝐴 and 𝐵 such that A ⊊ B , which are equidecomposable.

Proof

Suppose that 𝐺 has two equidecomposable subsets 𝐴 and 𝐵 such that A ⊊ B . In other words, there exist partitions { A 1 , … , A n } and { B 1 , … , B n } of 𝐴 and 𝐵 respectively and elements g 1 , … , g n of 𝐺 such that

Let 𝐻 be the subgroup generated by the finite set { g 1 , … , g n , x } , x ∈ B ∖ A . It is easy to see that A 1 := A ∩ H and B 1 := B ∩ H are equidecomposable and A 1 ⊊ B 1 . Consequently, 𝐻 is not finite by Lemma 2.4. This implies that 𝐺 is not locally finite. Suppose conversely that 𝐺 is not locally finite, and let 𝐻 be one of its infinite finitely generated subgroups. Then, again by Lemma 3.3, 𝐻 admits a subset 𝐵 equidecomposable with a proper subset 𝐴 of itself. Clearly, these sets are also equidecomposable subsets of the whole group 𝐺. This completes the proof.∎

Let 𝐺 be a finitely generated discrete group. Then the following conditions are equivalent.

1. 𝐺 is finite.

2. Every configuration equation Eq ⁡ ( g , E ) of 𝐺 admits a strictly positive solution.

Proof

Let 𝐺 be finite, and let ( g , E ) be a configuration pair for 𝐺. Defining

for every C ∈ Con ⁡ ( g , E ) , one achieves a strictly positive solution for Eq ⁡ ( g , E ) . Suppose on the other hand that 𝐺 is not finite. Then, by Lemma 3.3, there are two equidecomposable subsets 𝐴 and 𝐵 of 𝐺 such that A ⊊ B . In other words, D := B ∖ A ≠ ∅ , and there are partitions { A 1 , … , A n } and { B 1 , … , B n } of 𝐴 and 𝐵 respectively and elements g 1 , … , g n of 𝐺 such that B i = g i ⁢ A i ( 1 ≤ i ≤ n ). Set E = { A 1 , … , A n , D , B c } and g = ( g 1 - 1 , … , g n - 1 ) . Suppose that Eq ⁡ ( g , E ) has a strictly positive solution ( f C ) C ∈ Con ⁡ ( g , E ) . Then

and the first equation holds because ( f C ) C ∈ Con ⁡ ( g , E ) is a solution to the system Eq ⁡ ( g , E ) . This implies that ∑ x 0 ⁢ ( C ) ⊆ D f C = 0 , which contradicts the strict positivity of ( f C ) since D ≠ ∅ . ∎

Let 𝐺 be an infinite group. Then there exists a configuration pair ( g , E ) of 𝐺 such that Con ⁡ ( g , E ) ∉ Con ⁡ ( F ) for any finite group 𝐹.

Proof

It is clear by Theorem 3.5. ∎

Let 𝐺 be a discrete group. Then the following conditions are equivalent.

1. 𝐺 is locally finite.

2. Every configuration equation Eq ⁡ ( g , E ) of 𝐺 admits a strictly positive solution.

Proof

Let 𝐺 be a locally finite group, and let ( g , E ) be a configuration pair for 𝐺. We shall prove that Eq ⁡ ( g , E ) admits a strictly positive solution. Suppose that, for each C ∈ Con ⁡ ( g , E ) , x C is a fixed element of x 0 ⁢ ( C ) and 𝐻 is the group generated by the finite set g ∪ { x C : C ∈ Con ⁡ ( g , E ) } . Then, by assumption, 𝐻 is a finite group. Let E = { E 1 , … , E m } and g = { g 1 , … , g n } . For each 1 ≤ i ≤ m , set

Note that each F i is a non-empty set and ℱ is a partition for 𝐻. Consequently, ( g , F ) is a configuration pair for 𝐻. We prove that Con ⁡ ( g , E ) = Con ⁡ ( g , F ) . Let C = ( C 0 , C 1 , … , C n ) ∈ Con ⁡ ( g , E ) . Then

So we have

This means that

The inclusion Con ⁡ ( g , F ) ⊆ Con ⁡ ( g , E ) is clear since each F i ⊆ E i . Now, by Theorem 3.5, Eq ⁡ ( g , E ) = Eq ⁡ ( g , F ) admits a strictly positive solution.

For the converse, suppose that every configuration equation Eq ⁡ ( g , E ) of 𝐺 admits a strictly positive solution. Then, by Theorem 3.4 and an argument like that in the proof of Theorem 3.5, 𝐺 is a locally finite group. ∎

Let 𝐺 be a discrete group. Then 𝐺 is not locally finite if and only if it is paradoxical or it admits an amenable infinite finitely generated subgroup.

Theorem 3.9 (Stiemke’s theorem [5])

Exactly one of the following systems has a solution:

1. y T ⁢ A ⪈ 0 for some y ∈ R n ,

2. A ⁢ x = 0 , x > 0 , for some x ∈ R m .