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I’m having trouble with this exercise:

Let $G$ be the free group generated by $a$ and $b$. Prove that the commutator subgroup $G’$ is not finitely generated.

I found a suggestion that says to prove $G’$ is generated by the collection $\{[x^m,y^n]\mid m,n\in\mathbb{Z}\}$. I don’t know how to prove this, and how it helps.

- Finite set of zero-divisors implies finite ring
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If you could please provide me with some hints… Thanks!

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You can prove this using algebraic topology$^{\dagger}$: in fact, it follows from an exercise in Hatcher’s book (see this related math.SE question). Basically: use Reidemeister-Schreier. I have written out a “full-ish” proof – but if you are comfortable with covering spaces then this proof is ridiculously quick and elegant!

The covering-spaces proof that $F_2^{\prime}=[F(a, b), F(a, b)]\cong F_{\infty}$ is as follows: The free group $F(a, b)$ is the fundamental group of a bouquet of two circles:

Using $\widetilde{X}$ to denote the universal cover, by standard covering-space theory the derived subgroup $F_2^{\prime}=[F(a, b), F(a, b)]$ is isomorphic to the fundamental group of $\widetilde{X}/F_2^{\prime}$. Now, $\widetilde{X}$ is just the Cayley graph of $\langle a, b; -\rangle$,

while $\widetilde{X}/F_2^{\prime}$ is the Cayley graph of $\mathbb{Z}\times\mathbb{Z}$,

Then, $\pi_1(\widetilde{X}/F_2^{\prime})$ is the group with presentation $\langle V; R\cup T\rangle$ where $V$ is the set of $1$-cells (edges) of $\widetilde{X}/F_2^{\prime}$, $R$ is the set of $2$-cells of $\widetilde{X}/F_2^{\prime}$, and $T$ is a spanning tree for $\widetilde{X}/F_2^{\prime}$. Now, $\widetilde{X}/F_2^{\prime}$ has no $2$-cells, and taking any spanning tree we are still left with infinitely many edges (for example, take the tree $T=\bigcup_{y\in\mathbb{Z}}\{(x, y): x\in\mathbb{Z}\}\cup\{(0, x): x\in\mathbb{Z}\}$). Hence, we have an infinitely generated group with no relations, so $F_2^{\prime}=[F(a, b), F(a, b)]\cong F_{\infty}$ as required.

In fact, there is nothing special about $\mathbb{Z}\times\mathbb{Z}$ here:

**Thereom.** Let $G$ be an infinite group. If $\phi: F_n\rightarrow G$ is a surjection from the free group of rank $n$, $\infty>n>1$, to $G$ then $\ker(\phi)$ is either trivial or is not finitely generated.

*Proof*. Suppose $\ker(\phi)$ is non-trivial, and write $N:=\ker(\phi)$. Then, similar to the above, $\widetilde{X}/N$ is a graph; it is essentially the Cayley graph for $G$ given by the generating set $\operatorname{im}(\phi(X))\subset G$ (note the subtlety if there are $x_1, x_2\in X$ such that $\phi(x_1)=\phi(x_2)$). Now, as above, $N=\pi_1(\widetilde{X}/N)$ is the group with presentation $\langle V; R\cup T\rangle$ where $V$ is the set of $1$-cells (edges) of $\widetilde{X}/N$, $R$ is the set of $2$-cells of $\widetilde{X}/N$, and $T$ is a spanning tree for $\widetilde{X}/N$. As $\widetilde{X}/N$ is a graph it has no $2$-cells. It remains to show that for any spanning tree $T$ of $\widetilde{X}/N$ there are infinitely many edges $e$ in $(\widetilde{X}/N)\setminus T$. As $\ker(\phi)$ is non-trivial there exists a non-empty word $W\in F(X)$ such that $\phi(W)=1$. This corresponds to a loop in $\widetilde{X}/N$, and indeed at each vertex $v$ there exists such a loop $\mathcal{l}_{W, v}$ of length $d:=|W|$. Note that the tree $T$ cannot contain every edge of the loop $\mathcal{l}_{W, v}$. As $G$ is infinite there exists an infinite sequence of pairwise non-equal vertices $v_1, v_2, \ldots$ of $\widetilde{X}/N$ such that no two vertices $v_i, v_j$ are linked by a path of length $\leq2d$. Therefore, the loops $\mathcal{l}_{W, v_i}$ and $\mathcal{l}_{W, v_j}$ contain no edges in common. Hence, there are infinitely many edges $e$ in $(\widetilde{X}/N)\setminus T$. We therefore have an infinitely generated group with no relations, so $N$ is not finitely generated as required.

$^{\dagger}$ Serios recently reminded me of this proof in the comments to this fine math.SE answer.

Some ideas for you to work on:

An element of $\;F:=F(a,b)\;$ in normal form

$$\;w=a^{n_1}b^{m_1}\cdot\ldots\cdot a^{n_k}b^{m_k}\;,\;\;n_i,m_i\in\Bbb Z\;,\;\;n_1,m_k\in\Bbb Z$$

belongs to $\;F’\;$ iff the exponent sum of both letters is zero, meaning:

$$w\in F’\iff\sum n_i=\sum m_i=0$$

For example, $\;a^{-2}bab^{-1}a\in F’\;$ , but $\;a^{-1}ba^2b^{-1}\notin F’\;$ .

From the above, it is clear that $\;F’=\left\langle a^{-n}b^{-m}a^nb^m\;:\;\;{n\,,\,m\in\Bbb N}\right\rangle\;$ . To prove these are *free generators* you can try either to show that any normal word in those generators is the trivial element iff it is the void element (this seems to be a fairly non-so-hard approach), or perhaps the universal property of free groups.

Another argument based on algebraic topology applied to graphs:

**Property:** Let $F$ be a free group of finite rank $k$ and $H \lhd F$ be a nontrivial normal subgroup. If $H$ is finitely generated then it is a finite-index subgroup.

Now, it is possible to deduce that $D(\mathbb{F}_2)$ is not finitely-generated for instance by noticing that the quotient $\mathbb{F}_2/ D(\mathbb{F}_2)$ (namely the abelianization) is isomorphic to $\mathbb{Z}^2$ (and in particular is infinite).

**Proof.** $F$ may be viewed as the fundamental group of a bouquet $B$ of $k$ circles. Let $\hat{B} \to B$ be the covering associated to the subgroup $H$. Because $H \lhd F$, the covering is normal, that is the restriction of the action $F= \pi_1(B) \to \mathrm{Aut}(\hat{B})$ on the fiber over the base point is transitive. Since $B$ has only one vertex, we deduce that $\mathrm{Aut}(\hat{B})$ acts transitively on $\hat{B}$.

From now on, suppose by contradiction that $H$ is an infinite-index subgroup, that is $\hat{B}$ is infinite.

Because $H \neq \{1\}$, $\hat{B}$ contains a cycle $C$. Now let $v_0,v_1,\ldots \in \hat{B}$ be a sequence of vertices such that $v_0 \in C$ and $d(v_i,v_j) > 2 \mathrm{lg}(C)$, and let $\varphi_1, \varphi_2, \dots$ be a sequence of automorphisms of $\hat{B}$ such that $\varphi_{i+1}(v_i)=v_{i+1}$. Then $\varphi_1(C), \varphi_2(C),\dots$ is a sequence of disjoint cycles in $\hat{B}$ so that $H= \pi_1(\hat{B})$ cannot be finitely-generated. $\square$

**Nota Bene:** From Nielsen-Schreier formula, we know that there is an equivalence: a nontrivial normal subgroup is finitely-generated if and only if it is a finite-index subgroup.

More information may be found in the note Graphs and Free Groups and in references therein. The question When is $G \ast H$ solvable? is also relevant.

The answers given thus far are great, but no one has posted my favorite proof of this result yet. Consider the homomorphism $\phi\colon F_2\to \mathbb{Z}^2$ defined on the generators by $a\mapsto (1,0)$ and $b\mapsto (0,1)$. This is just the quotient map $F_2\to F_2/[F_2,F_2]$, but we’re going to reinterpret it in a slightly different way. For any $w\in F_2$,

$$ \phi(w)=\left(\#\{a\text{‘s in }w\}-\#\{\bar{a}\text{‘s in }w\},\#\{b\text{‘s in }w\}-\#\{\bar{b}\text{‘s in }w\}\right)$$

Don’t think of $\phi$ as the quotient by the commutator subgroup, just think of it as a funny map (you should check that it’s a homomorphism) defined on the Cayley graph of $F_2$ as described above. Let’s think about the $G=\ker \phi$ and suppose $G$ were finitely generated $G=\langle w_1,\ldots,w_k\rangle$. Think about the values $\phi$ takes on as you write out each $w_i$ in terms of $a$’s and $b$’s (or equivalently as you walk from the identity to $w_i$ in the Cayley graph). $\phi$ will increase, then decrease, then increase, then decrease, and so on until you reach $w_i$ where $\phi(w_i)=(0,0)$. The point is that in each coordinate of $\mathbb{Z}^2$, there will be some maximum value above which you never passed. This is true for each generator individually, so there is some big compact set in $\mathbb{Z}^2$ which you never leave when writing an individual generator.

Now write a word in your finite generating set of $G$, $w=w_{i_1}w_{i-2}\cdots w_{i_j}$, then write out that word in $a$’s and $b$’s and think about the values $\phi$ takes along the way. You will be stuck inside the same compact set in $\mathbb{Z}^2$ because each generator only makes it so far then returns to the origin.

To finish, you just say $\phi([a^n,b^n])\in G$, but to write this out in the Cayley graph of $F_2$, you must pass through vertices where $\phi$ is arbitrarily large. Thus, the finite set of $w_i$’s did not actually generate $G$.

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