For free groups, rank is a kind of dimension so it is surprising that a free group of rank can contain a free subgroup of rank ; it is even possible that be infinite! In this note, we propose three nice proofs of this fact:
Theorem: The free group of rank two has a subgroup isomorphic to the free group of countable rank.
In particular, we deduce that a free group of rank (possibely infinite) contains a subgroup isomorphic to for every !
1) A proof based on SQ-universality. A countable group is SQ-universal if every countable group is isomorphic to a subgroup of a quotient of . First, notice that a SQ-universal group has a subgroup isomorphic to .
Indeed, as a countable group, is isomorphic to a subgroup of a quotient ; let be a free basis of this subgroup. Let be the canonical surjection and, for all , let be such that . If is a reduced word, then since , so is a free basis of the subgroup , hence .
Therefore, it is sufficient to prove that is SQ-universal, that is (following Fred Calvin, Embeddings Countable Groups in 2-Generator Groups, The Amer. Math. Monthly, Vol. 100, No. 6, pp. 578-580):
Theorem: Every countable group is embeddable in a 2-generator group.
Proof. Let be a countable group. Notice first that is an injective morphism, so without loss of generality we may suppose that is a subgroup of .
Let be such that:
Let and for all . Then notice that
The other cases are left to the reader as exercice. Then defines an isomorphism between and . We conclude by noticing that is a subgroup of the -generator group .
Another proof can be found in our note Amalgamated products and HNN extensions (I): A theorem of B.H. Neumann.
2) A proof based on algebraic topology. The free group is the fundamental group of a bouquet of two circles . Therefore, if denotes the universal covering of , then the derived subgroup is the fundamental group of .
But is the usual Cayley graph of (Figure 1) and is the usual Cayley graph of (Figure 2):
If is a maximal subtree of , then is a free basis of , hence .
Explicitely, taking , we find
as a free basis of . More information on algebraic topology can be found in Hatcher’s book, Algebraic Topology (available on the author’s webpage).
3) A proof based on combinatorial group theory. Following Serre’s book Trees, we prove the following lemma (proposition 4 page 6):
Lemma: Let be two groups. For convenience, let . Then is a free basis of the derived subgroup of .
Proof. The derived subgroup of is the smallest subgroup so that be abelian. Therefore, we may deduce that is generated by . To prove that is a free basis, we prove by induction that the reduced word (over )
with , satisfies the two following conditions
- The length of (as an element of ) is at least ,
- The reduction of (as an element of ) finishes by .
The case is obvious. From now on, suppose it is true for and let
be a reduced word over . By induction hypothesis, we can write
where is reduced (as an element of ) and . If , there is no cancellation is the expression above, so . If , two cases happen:
- Case 1: . Then, setting , the word
is reduced (as an element of ), hence .
- Case 2: . Because is reduced and , necessarily , so setting ,
is reduced (as an element of ) hence .
Corollary: The derived subgroup of is isomorphic to .
Just for fun, notice the following consequence of our lemma:
Corollary: The only non-trivial soluble free product is .
Sketch of proof. Show the following straightforward results:
- A subgroup of a soluble group is soluble,
- A quotient of a soluble groupe is soluble,
- Deduce from 2 that a free group of rank at least two is not soluble,
- Deduce from 1 and 3 that a soluble group has no subgroup isomorphic to a free group of rank at least two,
- Conclude using our lemma.