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:

and

.

Let and for all . Then notice that

.

For example,

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.