Every torsion-free group is a subgroup of a torsion-free group with two conjugacy classes

From Groupprops
Jump to: navigation, search
This article gives the statement, and possibly proof, of an embeddability theorem: a result that states that any group of a certain kind can be embedded in a group of a more restricted kind.
View a complete list of embeddability theorems

Statement

Suppose G is a torsion-free group. Then, there exists a torsion-free group with two conjugacy classes L that contains G. Further, if G is countable, we can choose L to be countable.

Related facts

Generalizations

Facts used

  1. Isomorphic iff potentially conjugate: Suppose \sigma_i:H_i \to K_i, i \in I is a collection of isomorphisms between subgroups of G. Then, there exists a group G_1 containing G, with elements g_i \in G_1, such that conjugation by g_i, restricted to H_i, is equal to the isomorphism \sigma_i. Moreover, if G is torsion-free, we can choose G_1 to be torsion-free.

Proof

Given: A torsion-free group G.

To prove: There exists a torsion-free group L containing G such that L has two conjugacy classes.

Proof:

  1. There exists a torsion-free group G_1 containing G such that any two non-identity elements of G are conjugate in G_1: Let I be the set of all pairs of non-identity elements in G, and \sigma_i:H_i \to K_i be the isomorphism between the cyclic subgroups generated by the first element of the pair and the cyclic subgroup generated by the second element of the pair (sending the generator to the generator). Note that such an isomorphism exists because all non-identity elements have the same order (infinite). By fact (1), there exists a group G_1 satisfying the required conditions.
  2. There exists a chain G = G_0 \le G_1 \le G_2 \le \dots where any two non-identity elements of G_{i-1} are conjugate in G_i: This follows by repeating the previous step.
  3. Define L as the ascending union (or direct limit) of the G_is. Then, any two non-identity elements of L are conjugate in L.