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:

$(m,n,p)a=(m+1,n,p)$

and

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

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):

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. $\square$