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The Magnus representation of a free group

In this post we will explain the Magnus embedding of a free. This gives a faithful representation of a non-Abelian free group as a (multiplicative) group of power series in multiple non-commuting variables which has many nice consequences on the group-theoretic properties of the free group.


1. Power series in non-commuting variables

Let \(X\) be a set. The Magnus ring in the variables \(X\) is the ring \(\mathbb Z\langle\!\langle X \rangle\!\rangle\) of infinite series in the freely non-commuting variables \(X\). We remark that, for us, rings are always unital, so the power \(1\) is an element of \(\mathbb Z\langle\!\langle X \rangle\!\rangle\). For an expression to count as a power series, it must have finitely many terms in each degree, where the elements of \(X\) have degree one. For instance, suppose that \(X\) is the infinite set \(\{x_1, x_2, x_3, \dots\}\). Then \[1 + x_1 + x_1^2 + x_1^3 + \cdots\] is a power series, while \[x_1 + x_2 + x_3 + \cdots\] is not. A more formal defintion of \(\mathbb Z\langle\!\langle X \rangle\!\rangle\) can be given as follows. Let \(\mathbb Z\langle X\rangle\) be the free (non-commutative) unital \(\mathbb Z\)-algebra generated by \(X\) and let \(\mathfrak a_X\) be the two-sided ideal of \(\mathbb Z\langle X \rangle\) generated by \(X\). Then \[ \mathbb Z\langle\!\langle X \rangle\!\rangle = \varprojlim_{n \in \mathbb N} \ \mathbb Z\langle X\rangle / \mathfrak a_X^n. \]

Let \(U_X\) be the unit group of \(\mathbb Z\langle\!\langle X \rangle\!\rangle\) and let \(\omega_X\) denote the two-sided ideal of \(\mathbb Z\langle\!\langle X \rangle\!\rangle\) generated by \(X\). As the following lemma shows, \(U_X\) is a very large group, which is something we will exploit shortly.

Proof. The expression \(1 - f + f^2 - f^3 + \cdots \) defines a valid element of \(\mathbb Z\langle\!\langle X \rangle\!\rangle\). This is because for each \(n \in \mathbb N\), the element \(f^i\) can contain terms of degree \(n\) only if \(i \leqslant n\). A direct computation then shows that \[(1 + f)^{-1} = 1 - f + f^2 - f^3 + \cdots \tag*{\(\square\)} \]

2. The Magnus embedding

Let \(S = \{s_x : x \in X\}\) be a set in one-to-one correspondence with \(X\) and let \(F(S)\) be the (non-Abelian) free group freely generated by \(S\). The universal property of the free group states that given any group \(G\) and any set theoretic map \(S \to G\), there exists a unique group homomorphism \(F(S) \to G\) extending the set-theoretic map. We can now define the Magnus map, the main subject of this post.

Definition. The Magnus map is the group homomorphism \(\mu_X \colon F(S) \to U_X\) determined by the assignment \(s_x \mapsto 1 + x\) for each \(x \in X\).

By Lemma 1, \(\mu_X\) is well defined. The Magnus map will be used to study the free group; for this we need to establish the following fundamental fact.

Proof. Let \(w = s_{x_1}^{n_1} \cdots s_{x_m}^{n_m}\) be a non-identity reduced word in \(F(S)\), meaning that \(x_i \neq x_{i+1}\) and \(n_i \neq 0\) for all \(i\). By definition, \[ \mu_X(w) = (1 + x_1)^{n_1} \cdots (1+x_m)^{n_m}. \] By a direct calculation, one checks that the monomial \(x_1 \cdots x_m\) appears with non-zero coefficient \(n_1 \cdots n_m\). Therefore, \(\mu_X(w) \neq 1\), which proves that \(\mu_X\) is injective. \(\square\)

3. Consequences

Given a group \(G\), and two subgroups \(A, B \leqslant G\), we denote by \([A,B]\) the subgroup of \(G\) generated by all the commutators \([a,b] = a^{-1}b^{-1}ab\) with \(a \in A\) and \(b \in B\).

A group \(G\) is residually nilpotent if it admits a descending chain of subgroups \(G = G_1 \geqslant G_2 \geqslant \dots\) such that \([G,G_n] \leqslant G_{n+1}\) for each \(in \in \mathbb N\) and moreover \(\bigcap_{n \in \mathbb N} G_n = \{1\}\). If \(G_n = \{1\}\) for some \(n\), then \(G\) is nilpotent. If \(G\) admits a chain \((G_n)_{n \in \mathbb N}\) witnessing the residual nilpotence property, and moreover \(G/G_n\) is torsion-free for all \(n\), then \(G\) is residually torsion-free nilpotent.

Proof. Let \(G = 1 + \omega_X\) and let \(G_n = 1 + \omega_X^n\) for all \(n \in \mathbb N\). We will show that the chain \((G_n)_{n\in \mathbb N}\) witnessing the residual torsion-free nilpotence of \(G\). Let \(g \in G\) be a non-trivial element. Then there is an element \(f \in \omega_X\) such that \(g = 1 + f\). If the minimal degree term of \(f\) is of degree \(n\), then \(f \notin \omega_X^{n+1}\), and therefore \(g \notin 1 + \omega_X^{n+1}\). Hence, \(\bigcap_n G_n = \{1\}\).

Fix some \(n \in \mathbb N\) and let \(f \in \omega_X^n\) and \(h \in \omega_X\). Then \[ (1 + f)^{-1}(1 + h)^{-1}(1 + f)(1 + h) = (1 - f + f^2 + \cdots)(1 - h + h^2 + \cdots)(1 + f)(1 + h) \] from which ones sees directly that there are no terms of degree at most \(n\) in the product, except for \(1\). Hence, \([G,G_n] \leqslant G_{n+1}\). This shows that \(G\) is residually nilpotent.

Let \(f \in \omega_X^n \smallsetminus \omega_X^{n+1}\). Then \((1 + f)^n = 1 + nf + \cdots\), where the dots are placeholders for terms of degree at least \(n+1\). Hence, \((1 + f)^n \notin 1 + \omega_X^{n+1}\) for all \(n \neq 0\), proving that \(G/G_n\) is torsion-free. Hence, \(G\) is residually torsion-free nilpotent. \(\square\)

Proof. We first prove that subgroups of residually torsion-free nilpotent groups are residually torsion-free nilpotent. Suppose \(G\) is a residually torsion-free nilpotent group and \(H \leqslant G\) is a subgroup. If \((G_n)_{n\in\mathbb N}\) is a chain of subgroups witnessing the fact that \(G\) is residually torsion-free nilpotent, then the chain \((H \cap G_n)_{n\in\mathbb N}\) witnesses the residual torsion-free nilpotence of \(H\). By Theorem 2, the free group \(F(S)\) is isomorphic to a subgroup of \(1 + \omega_X\) and therefore is itself residually torsion-free nilpotent by Theorem 3. Since \(S\) is arbitrary, we conclude that free groups are residually torsion-free nilpotent. \(\square\)

By a theorem of Gruenberg, finitely generated torsion-free nilpotent groups are residually \(p\)-finite for all primes \(p\) (meaning that every non-trivial element survives in a finite \(p\)-quotient). Combining this with the previous corollary yields that free groups are residually \(p\)-finite for all primes \(p\). We will give a proof of this fact without appealing to Gruenberg's theorem. It is also possible to give geometric proofs of the residual \(p\)-finiteness of free groups using covering space theory, but we will not discuss that here.

Proof. First observe that it suffices to prove this for finitely generated free groups. Indeed, given any non-trivial element \(w \in F(S)\), there is a finite set of letters \(S_0 \subseteq S\) such that \(w \in F(S_0) \subseteq F(S)\) and moreover there is a retraction \(F(S) \to F(S_0)\). This shows that free groups are residually (finitely generated free), which reduces the proof to the case where \(S\) is finite.

Let \(\mathbb Z/p^k \langle X \rangle\) denote the free unital \(\mathbb Z/p^k\)-algebra on the (finitely many) variables \(X\), and let \(\omega_{X,p}\) be the two-sided ideal generated by \(X\). The ring \( (\mathbb Z/p^k \langle X \rangle)/\omega_{X,p}^n \) is then a finite ring, consisting of polynomials over \(\mathbb Z/p^k\) truncated at degree \(n\). By the binomial theorem, the image of \(1 + \omega_X\) under the composition \[ \mathbb Z\langle\!\langle X \rangle\!\rangle \to \mathbb Z\langle\!\langle X \rangle\!\rangle/\omega_X \to (\mathbb Z/p^k \langle X \rangle)/\omega_{X,p}^n \] is a finite \(p\)-group (viewed as a multiplicative group), where the second map is reduction of coefficients modulo \(p^k\). It is easy to see that for every element \(f \in \mathbb Z\langle\!\langle X \rangle\!\rangle\), there are integers \(n\) and \(k\) such that \(f\) survives under the above composition. It thus follows that \(1 + \omega_X\) is residually \(p\)-finite. But subgroups of residually \(p\)-finite groups are residually \(p\)-finite, so \(F(S)\) is residually \(p\)-finite by Theorem 2. \(\square\)

Next we study orderability of the free group. A group \(G\) is orderable if there is a total order \(\prec\) on \(G\) such that if \(g,h \in G\) are such that \(g \prec h\), then \(tg \prec th\) and \(gt \prec ht\) for all \(t \in G\). It is clear that orderability is a property that passes to subgroups, so in order to prove that free groups are orderable, it will suffice to show that \(1 + \omega_X\) is orderable.

Part of the difficulty in ordering the free group is that there is no natural invariant lexicographic ordering on the elements of \(F(S)\) induced by a total order on \(S\). This comes from the fact that for every \(s \in S\), there is an inverse element \(s^{-1} \in S\), but this problem goes away in \(\mathbb Z\langle\!\langle X \rangle\!\rangle\), since there all monomials are "positive words" in the letters \(X\).

Proof. As discussed above, we will prove that \(1 + \omega_X\) is orderable, from which the theorem follows by Theorem 2. Put an arbitrary total order \(\prec_m\) on the elements of \(X\). We extend \(\prec_m\) to all monomials as follows: let \(m_1\) and \(m_2\) be monomials in the letters \(X\). If the degree of \(m_1\) is less than the degree of \(m_2\), we declare \(m_1 \prec_m m_2\). Now suppose \(m_1\) and \(m_2\) have equal degree \(n\) and write \(m_1 = x_1 x_2 \cdots x_n\) and \(m_2 = y_1 y_2 \cdots y_n\) where \(x_i, y_i \in X\) for all \(i\). If \(m_1 \neq m_2\), then there is a minimal index \(i\) such that \(x_i \neq y_i\). Put \(m_1 \prec_m m_2\) if and only if \(x_i \prec_m y_i\).

We are now ready to define the order on \(1 + \omega_X\). For this we will use the following definition: a subset \(P\) of a group \(G\) is a positive cone if

  1. \(P \sqcup \{1\} \sqcup P^{-1}\) forms a partition of \(G\),
  2. \(P\) is invariant under conjugation, and
  3. \(P \cdot P \subseteq P\).
If \(G\) admits a positive cone \(P\) then it is orderable with the following order: \(g \prec h\) if and only if \(g^{-1}h \in P\).

We will say that the leading coefficient of a non-trivial element \(1 + f \in 1 + \omega_X\) is the coefficient of the monomial of \(f\) that is minimal under the monomial order \(\prec_m\) defined in the first paragraph. Let \(P \subseteq 1 + \omega_X\) be the set of non-trivial elements \(1 + f\) with positive leading coefficient. We conclude by verifying that \(P\) satisfies the properties (i), (ii), and (iii) above.

Property (i) follows from the fact that \( (1 + f)^{-1} = 1 - f + f^2 - \cdots\). So if the leading coefficient of \(1 + f\) is positive, then that of \((1 + f)^{-1}\) is negative, and vice versa. We check (ii): let \(1 + f \in P\) and let \(1 + g \in 1 + \omega_X\). Then \begin{align*} (1+g)^{-1}(1+f)(1+g) &= 1 + (1+g)^{-1}f(1+g) \\ &= 1 + \sum_{n=0}^\infty (-g)^n f(1 + g) \\ &= 1 + f + fg + \sum_{n=1}^\infty (-g)^n f(1 + g) \end{align*} and one sees that the leading coefficient of \((1+g)^{-1}(1+f)(1+g)\) is the same as that of \(1+f\). This verifies (ii). We leave (iii) as an exercise, as it is also a direct computation. \(\square\)

Given a ring \(R\) and a group \(G\), the group ring \(R[G]\) is the ring of finite linear combintaions \(\sum_{g \in G} r_g g\) with \(r_g \in G\) for all \(g \in G\). We will give positive resolutions to all of the following conjectures for group algebras of free groups, using the biorderability of free groups.

The Kaplansky Conjectures. Let \(\mathbb K\) be a field and let \(G\) be a torsion-free group. Then

  1. \(\mathbb K[G]\) has no non-trivial units, i.e. all its units are of the form \(\lambda g\) for some \(\lambda \in \mathbb K \smallsetminus \{0\}\) and \(g \in G\);
  2. \(\mathbb K[G]\) is a domain;
  3. \(\mathbb K[G]\) as no non-trivial idempotents, i.e. if \(x^2 = x\), then either \(x = 1\) or \(x = 0\);
  4. \(\mathbb K[G]\) is directly finite, i.e. if \(xy = 1\), then \(yx = 1\).

For any group algebra \(\mathbb K[G]\), there are implications (i) \(\Rightarrow\) (ii) \(\Rightarrow\) (iii) \(\Rightarrow\) (iv). The only non-obvious implication is (i) \(\Rightarrow\) (ii). There is a strengthening of (iv), which does not follow from the other Kaplansky Conjectures.

The Stable Finiteness Conjecture. Let \(G\) be any group. Then \(\mathbb K[G]\) is stably finite, meaning that if \(A\) and \(B\) are any \(n \times n\) matrices over \(\mathbb K[G]\) such that \(AB\) is the identity matrix, then \(BA\) is the identity matrix.

Proof. We begin by proving the first of the Kaplansky Conjectures on the structure of units in \(\mathbb K[F]\). Let \(a =\alpha_1 g_1 + \cdots + \alpha_m g_m\) and \(b = \beta_1 h_1 + \cdots + \beta_n h_n\) be two elements of \(\mathbb K[F]\) such that \(ab = 1\), and moreover suppose that \(\alpha_i \neq 0\) and \(\beta_i \neq 0\) for all appropriate \(i\). By Theorem 6, there exists a total bi-invariant order \(\prec\) on \(F\). Without loss of generality, we can also assume that \(g_1 \preccurlyeq \cdots \preccurlyeq g_m\) and \(h_1 \preccurlyeq \cdots \preccurlyeq h_n\). Suppose that \(g_1 \neq g_m\). Then \(g_1 h_1\) and \(g_m h_n\) both appear in the product \(ab\) with non-zero coefficients \(\alpha_1\beta_1\) and \(\alpha_m\beta_n\) respectively (note we are using the invariance of \(\prec\) to conclude that \(g_1h_1 \neq g_mh_n\)). But this contradicts \(ab = 1\), so we must have \(m = 1\). This proves Kaplansky Conjecture (i).

We have already mentioned that (i) \(\Rightarrow\) (ii) \(\Rightarrow\) (iii) \(\Rightarrow\) (iv), but that the first implication (i) \(\Rightarrow\) (ii) is non-trivial. Thankfully, a very similar argument to the one in the paragraph above yields (ii) as well; we leave this to the reader. \(\square\)

One can prove something stronger than (ii), which is that \(\mathbb K[F]\) embeds into a division ring. We note that the following is also open.

Conjecture. Let \(\mathbb K\) be a field and let \(G\) be a torsion-free group. Then \(\mathbb K[G]\) embeds into a division ring.

The following result holds for any orderable group. The proof, which we only sketch, is due to Malcev and Neumann, independently.

Proof (sketch). By Theorem 6, there is a total bi-invariant order \(\prec\) on \(F\). Consider the following set of formal power set of formal power series: \[ \mathcal{MN}_\prec(\mathbb K[F]) = \left\{ \sum_{g\in G} \alpha_g g : \{g : \alpha_g \neq 0\} \ \text{is well ordered under} \prec \right\}. \] We have that \(\mathbb K[F] \subseteq \mathcal{MN}_\prec(\mathbb K[F])\) in a natural way. One can show that the operations of addition and multiplication on \(\mathbb K[F]\) naturally extend to \(\mathcal{MN}_\prec(\mathbb K[F])\); this reduces to showing that if \(A\) and \(B\) are well-ordered subset of \(\mathbb K[F]\), then \(A \cup B\) and \(AB\) are well-ordered. This makes \(\mathcal{MN}_\prec(\mathbb K[F])\) into a ring called the Malcev–Neumann power series ring of \(\mathbb K[F]\) associated to the order \(\prec\). The key point is that \(\mathcal{MN}_\prec(\mathbb K[F])\) is actually a division ring. To see this, let \(x \in \mathcal{MN}_\prec(\mathbb K[F])\) be an arbitrary non-zero element. By multiplying \(x\) by some unit of the form \(\alpha g\) for some \(\alpha \in \mathbb K\) and \(g \in G\), we can assume that \(x = 1 + y\) where every element in the support of \(y\) is greater than \(1\) with respect to \(\prec\). Then \(x^{-1} = 1 - y + y^2 - y^3 + \cdots\) (we leave it as an exercise to check this is a well defined element of \(\mathcal{MN}_\prec(\mathbb K[F])\)). \(\square\)

As a corollary, we can also establish the Stable Finiteness Conjecture for group algebras of free groups.

Proof. By Theorem 8, \(\mathbb K[F]\) embeds into a division ring. Since division rings are stably finite and stable finiteness passes to subrings, we conclude that \(\mathbb K[F]\) is stably finite. \(\square\)