For free groups, rank is a kind of dimension so it is surprising that a free group of rank n can contain a free subgroup of rank m>n; it is even possible that m be infinite! In this note, we propose three nice proofs of this fact:

Theorem: The free group \mathbb{F}_2 of rank two has a subgroup isomorphic to the free group \mathbb{F}_{\infty} of countable rank.

In particular, we deduce that a free group of rank n \geq 2 (possibely infinite) contains a subgroup isomorphic to \mathbb{F}_m for every m \leq \aleph_0!

1) A proof based on SQ-universality. A countable group G is SQ-universal if every countable group is isomorphic to a subgroup of a quotient of G. First, notice that a SQ-universal group G has a subgroup isomorphic to \mathbb{F}_{\infty}.

Indeed, as a countable group, \mathbb{F}_{\infty} is isomorphic to a subgroup of a quotient G/H; let \{x_1,x_2, \dots\} be a free basis of this subgroup. Let \pi : G \to G/H be the canonical surjection and, for all i \geq 1, let x_i^* \in G be such that \pi(x_i^*)=x_i. If w is a reduced word, then w(x_1^*,x_2^*,\dots) \neq 1 since \pi(w(x_1^*,x_2^*,\dots))=w(x_1,x_2,\dots) \neq 1, so \{x_1^*,x_2^*,\dots\} is a free basis of the subgroup F = \langle x_1^*,x_2^*,\dots \rangle \leq G, hence F \simeq \mathbb{F}_{\infty}.

Therefore, it is sufficient to prove that \mathbb{F}_2 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 G= \{g_1,g_3,g_5, \dots \} be a countable group. Notice first that \varphi : \left\{ \begin{array}{ccc} G & \to & \mathrm{Sym}(G) \\ g & \mapsto & (h \mapsto gh) \end{array} \right. is an injective morphism, so without loss of generality we may suppose that G is a subgroup of \mathrm{Sym}(\mathbb{N}) \simeq \mathrm{Sym}(G).

Let a,b \in \mathrm{Sym}(\mathbb{Z} \times \mathbb{Z} \times \mathbb{N}) be such that:



(m,n,p)b = \left\{ \begin{array}{cl} (m,n+1,p) & \text{if} \ m=0 \\ (m,n,g_m \cdot p) & \text{if} \ m \ \text{is odd}, m>0,n \geq 0 \\ (m,n,p) & \text{otherwise} \end{array} \right..

Let b_i=a^iba^{-i} and \hat{g}_i=b_ib^{-1}b_i^{-1}b for all i=1,3,5, \dots. Then notice that

(m,n,p) \hat{g}_i = \left\{ \begin{array}{cl} (0,0,g_i \cdot p) & \text{if} \ m=n=0 \\ (m,n,p) & \text{otherwise} \end{array} \right..

For example,

\begin{array}{ll} (0,0,p) \hat{g}_i & = (0,0,p) a^iba^{-i}b^{-1}b_i^{-1}b = (i,0,p)ba^{-i}b^{-1}b_i^{-1}b \\ & = (i,0, g_i \cdot p)a^{-i}b^{-1}b_i^{-1}b = (0,0,g_i \cdot p) b^{-1}b_i^{-1}b \\ & = (0,-1,g_i \cdot p) a^ib^{-1} a^{-i} b = (i,-1,g_i \cdot p)b^{-1}a^{-i}b \\ & = (i,-1,g_i \cdot p) a^{-i}b = (0,-1,g_i \cdot p) b =(0,0,g_i \cdot p) \end{array}

The other cases are left to the reader as exercice. Then g_i \mapsto \hat{g}_i defines an isomorphism between G and \hat{G} = \{\hat{g}_1,\hat{g}_3,\hat{g}_5, \dots\}. We conclude by noticing that \hat{G} is a subgroup of the 2-generator group \langle a,b \rangle \leq \mathrm{Sym}(\mathbb{Z} \times \mathbb{Z} \times \mathbb{N}). \square

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 \mathbb{F}_2 is the fundamental group of a bouquet of two circles X. Therefore, if \hat{X} denotes the universal covering of X, then the derived subgroup D := [\mathbb{F}_2,\mathbb{F}_2] is the fundamental group of \hat{X}/D.

But \hat{X} is the usual Cayley graph of \mathbb{F}_2 (Figure 1) and \hat{X} /D is the usual Cayley graph of \mathbb{Z}^2 (Figure 2):

Cayley graphs

If T is a maximal subtree of \hat{X}/D, then \{ \text{edges of} \ \hat{X}/D \ \text{not in} \ T\} is a free basis of D, hence D \simeq \mathbb{F}_{\infty}. \square

Explicitely, taking T = (\{0\} \times \mathbb{R}) \cup \bigcup\limits_{i \in \mathbb{Z}} \mathbb{R} \times \{i\}, we find

\{a^ib^ja^{-1}b^{-j}a^{1-i},a^{-i}b^jab^{-j}a^{i-1} \mid i>0,j \in \mathbb{Z} \backslash \{0\} \}

as a free basis of D. 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 A,B be two groups. For convenience, let S= \{[a,b] \mid a \in A \backslash \{1\}, b \in B \backslash \{1\} \}. Then S is a free basis of the derived subgroup of A \ast B.

Proof. The derived subgroup D of A \ast B is the smallest subgroup so that (A \ast B) / D be abelian. Therefore, we may deduce that D is generated by S. To prove that S is a free basis, we prove by induction that the reduced word (over S)

g = [a_1,b_2]^{\epsilon_1} \cdots [a_n,b_n]^{\epsilon_n}

with \epsilon_i \in \{ \pm 1\}, satisfies the two following conditions

  1. The length of g (as an element of A \ast B) is at least n+3,
  2. The reduction of g (as an element of A \ast B) finishes by a_n^{-1}b_n^{-1}.

The case n=1 is obvious. From now on, suppose it is true for n and let

g = [a_1,b_1]^{\epsilon_1} \cdots [a_{n+1},b_{n+1}]^{\epsilon_{n+1}} \hspace{1cm} (1)

be a reduced word over S. By induction hypothesis, we can write

g = s_1 \cdots s_pa_n^{-1} b_n^{-1} [a_{n+1},b_{n+1}]^{\epsilon_{n+1}},

where s_1 \cdots s_pa_n^{-1} b_n^{-1} is reduced (as an element of A \ast B) and p \geq n+1. If \epsilon_{n+1}=1, there is no cancellation is the expression above, so \ell g(g) = p+6 \geq n+7. If \epsilon_{n+1}=-1, two cases happen:

  • Case 1: b_n \neq b_{n+1}. Then, setting \tilde{b}= b_n^{-1}b_{n+1}, the word

g = s_1 \cdots s_p a_n^{-1} \tilde{b} a_{n+1}b_{n+1}^{-1}a_{n+1}^{-1}

is reduced (as an element of A \ast B), hence \ell g(g) = p+5 \geq n+6.

  • Case 2: b_n=b_{n+1}. Because (1) is reduced and \epsilon_{n+1}=-\epsilon_n, necessarily a_n \neq a_{n+1}, so setting \tilde{a} = a_n^{-1} a_{n+1},

g= s_1 \cdots s_p \tilde{a} b_{n+1}^{-1} a_{n+1}^{-1}

is reduced (as an element of A \ast B) hence \ell g(g) = p+3 \geq n+4. \square

Corollary: The derived subgroup of \mathbb{F}_2 is isomorphic to \mathbb{F}_{\infty}.

Just for fun, notice the following consequence of our lemma:

Corollary: The only non-trivial soluble free product is \mathbb{Z}_2 \ast \mathbb{Z}_2.

Sketch of proof. Show the following straightforward results:

  1. A subgroup of a soluble group is soluble,
  2. A quotient of a soluble groupe is soluble,
  3. Deduce from 2 that a free group of rank at least two is not soluble,
  4. Deduce from 1 and 3 that a soluble group has no subgroup isomorphic to a free group of rank at least two,
  5. Conclude using our lemma. \square