$\mathrm{GL}_n(\mathbb{Z}_2)=\mathrm{Out}(F_n)/\langle\langle \epsilon_1,\dots,\epsilon_n\rangle\rangle$

Question

In a paper I found the following result: $$\mathrm{GL}_n(\mathbb{Z}_2)=\mathrm{Out}(F_n)/\langle\langle \epsilon_1,\dots,\epsilon_n\rangle\rangle$$ However, they got the result as a corollary of a pretty strong result. I was wondering if there is a direct proof for this fact. Here $\epsilon_i$ is the automorphism of $F_n$ that inverts the $i$-th generator and leaves the rest fixed.

Thanks for your help.

Edit: The natural approach I was thinking is to consider the map given by the abelianization and compose it with the reduction mod $2$: $$\varphi:\mathrm{Out}(F_n)\to\mathrm{GL}_n(\mathbb{Z})\to\mathrm{GL}_n(\mathbb{Z}_2)$$ This is trivially an epimorphism and I proving that $\ker(\varphi)=\langle\langle \epsilon_1,\dots,\epsilon_n\rangle\rangle$ will solve the problem. $\mathrm{Out}(F_n)$ is generated by $\epsilon_1,\dots,\epsilon_n,\lambda_{i,j}$ with $i,j=1,\dots,n$ and $i\neq j$. ($\lambda_{i,j}(v_i)=v_iv_j$ and $\lambda_{i,j}(v_k)=v_k$ if $k\neq i$, where $v_1,\cdots,v_n$ are the generators of $F_n$). But there are elements such as $\lambda_{i,j}^2$ that lie in $\ker(\varphi)$ and I cannot prove that they lie in $\langle\langle \epsilon_1,\dots,\epsilon_n\rangle\rangle$.

Answer 1

$\DeclareMathOperator{Out}{Out} \DeclareMathOperator{Aut}{Aut} \DeclareMathOperator{IA}{IA}$ The key thing that is missing is that the Torelli subgroup of $\Out(F_n)$ is contained in the normal closure of these elements. This is in fact true, and is a very special case of a theorem will appear in my student Jiayi Shen's thesis. More generally, her thesis will prove that in most cases, the $\Out(F_n)$-normal closure of a finite subgroup of $\Out(F_n)$ is what you would guess to be (there are a few cases, and more generally she has to assume that $n \geq 3$ and can't handle everything that happens at the prime $2$, though for the special case asked here she has no issues). Since it is not yet posted to the arXiv, I will give the argument here specialized to this particularly easy case.

Instead of $\Out(F_n)$, I will prove it for $\Aut(F_n)$ (which is enough). Let $\IA_n$ be the Torelli subgroup of $\Aut(F_n)$. Let $\{x_1,\ldots,x_n\}$ be the standard basis of $F_n$. For distinct $1 \leq i,j \leq n$, let $C_{ij} \in \IA_n$ be the element that conjugates $x_i$ by $x_j$. A classical theorem of Nielsen says that the $\Aut(F_n)$-normal closure of $C_{12}$ is $\IA_n$, so it is enough to prove that $C_{12}$ is in the indicated subgroup.

One element of the subgroup in question is the map $f\colon F_n \rightarrow F_n$ that inverts $x_1$ and $x_2$ and fixes all the other $x_i$. Let $L_{ij} \in \Aut(F_n)$ be the Nielsen transformation that multiplies $x_i$ on the left by $x_j$. The subgroup in question then contains $$(L_{12} f L_{12}^{-1}) f^{-1},$$ which equals $C_{12}$ by the following calculation:

1. First,

$$L_{12} f L_{12}^{-1} f^{-1}(x_1) = L_{12} f L_{12}^{-1}(x_1^{-1}) = L_{12} f (x_1^{-1} x_2) = L_{12}(x_1 x_2^{-1}) = x_2 x_1 x_2^{-1}.$$

2. Second, for $i \geq 2$ for an appropriate choice of sign depending on whether $i=2$ or $i \geq 3$ we have

$$L_{12} f L_{12}^{-1} f^{-1}(x_i) = L_{12} f L_{12}^{-1}(x_i^{\pm 1}) = L_{12} f(x_i^{\pm 1}) = L_{12}(x_i) = x_i.$$

Answer 2

Edit. Assertion $(ii)$ of the Claim below is now given by the commutator computation of Andy Putman's answer (see Remark 2 below for notational match). This provides us with a proof of OP's isomorphism.

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The isomorphism result mentioned in OP's question, that is [3, Proposition 2.8], holds true if $n = 2$. The general case can be reduced to the following statement:

Claim. Let $n \ge 2$. Let $F_n$ be the free group with basis $\{a_1, \dots, a_n\}$. Let $\operatorname{Inn}(F_n)$ the group of inner automorphisms of $F_n$. Let $\epsilon_i, \rho_{ij}$ and $\lambda_{ij}$ be defined as in [3] (see re-definitions below). Then the following are equivalent:

• $(i)$ The quotient group $Q:=\operatorname{Out}(F_n)/ \langle \langle \epsilon_1, \dots, \epsilon_n \rangle\rangle$ is isomorphic to $\operatorname{GL}_n(\mathbb{Z}/2\mathbb{Z})$.
• $(ii)$ The image of $\alpha_{112} := \rho_{12} \lambda_{12}^{-1}$ is trivial in $Q$, i.e., we have $$\alpha_{112} \in \langle \langle \operatorname{Inn}(F_n), \epsilon_1, \dots, \epsilon_n \rangle\rangle.$$

We adopt the following convention: $[x, y] = x^{-1}y^{-1}xy$.

Remark 1. We have $\alpha_{112}^2 \in \langle \langle \epsilon_1, \dots, \epsilon_n \rangle\rangle$ since $[\epsilon_1, \alpha_{112}] = \alpha_{112}^2$.

Remark 2. By Andy Putman's computation, we actually have $\alpha_{112} = [\lambda_{12}, \epsilon_1 \epsilon_2]$, thus assertion $(ii)$ is true.

Notation match. $L_{12} = \lambda_{12}, C_{12} = \alpha_{112}^{-1}$ and $f = \epsilon_1 \epsilon_2$.

In keeping with [3, Definition 2.4], the automorphisms $\epsilon_i, \rho_{ij}$ and $\lambda_{ij}$ of $F_n$ are defined as follows: for every $i, j, k \in \{1, \dots, n\}$ such that $i \notin \{j, k\}$, we set $$\rho_{ij}(a_i) = a_i a_j, \lambda_{ij}(a_i) = a_j a_i, \text{ and }\epsilon_i(a_i) = a_i^{-1}, $$ $$\rho_{ij}(a_k) = \lambda_{ij}(a_k) = \epsilon_i(a_k) = a_k.$$

We shall derive the claim from the next three lemmas. We will only show the third lemma as the others are well-known.

First, let us observe that since the derived subgroup of $F_n$ is characteristic, every automorphism $\phi$ of $F_n$ induces an automorphism $\overline{\phi}$ of $\mathbb{Z}^n$. The group of the automorphisms of $F_n$ which induces the identity on $\mathbb{Z}^n$ is the group $\operatorname{IA}(F_n)$ of the Bachmuth automorphisms of $F_n$. Clearly, $\operatorname{IA}(F_n)$ contains $\operatorname{Inn}(F_n)$. Nielsen showed that $\operatorname{IA}(F_n)$ is normally generated by $\alpha_{112}$ for every $n \ge 2$ [1, Remarks following Proposition I.4.15]. It is also known that $\operatorname{IA}(F_2) = \operatorname{Inn}(F_2)$, see [1, Proposition I.4.5] while the Torelli group $\operatorname{IA}(F_n)/\operatorname{Inn}(F_n)$ is non-trivial and torsion free if $n > 2$ [1, Corollary I.4.12]. Let $i, j, k \in \{1, \dots, n\}$ such that $k \neq j$. Define $\alpha_{ijk} \in \operatorname{Aut}(F_n)$ through $$ \alpha_{ijk}(a_i) = a_i[a_j, a_k], \quad \alpha_{ijk}(a_l) = a_l \text{ for every } l \neq i. $$ Although we won't need this fact, it's worth mentioning that Magnus and Nielsen showed that the automorphisms $\alpha_{ijk}$ generate $\operatorname{IA}(F_n)$ [1, Remarks following Proposition I.4.15].

Lemma 1.[1, Proposition I.4.4] Let $n \ge 1$. The group homomorphism $\phi \mapsto \overline{\phi}$ induces an isomorphism $\operatorname{Aut}(F_n)/\operatorname{IA}(F_n) \simeq \operatorname{GL}_n(\mathbb{Z})$.

Let $m \in \mathbb{N}$. We denote by $\operatorname{SL}_n(\mathbb{Z}, m\mathbb{Z})$ the kernel of the natural map $$\operatorname{SL}_n(\mathbb{Z}) \twoheadrightarrow \operatorname{SL}_n(\mathbb{Z}/m\mathbb{Z}).$$

We denote by $\operatorname{E}_n(\mathbb{Z})$ the subgroup of $\operatorname{SL}_n(\mathbb{Z})$ generated by the matrices $e_{ij} := \overline{\lambda_{ij}}$. We denote by $\operatorname{E}_n(\mathbb{Z}, m\mathbb{Z})$ the normal subgroup of $\operatorname{E}_n(\mathbb{Z})$ normally generated by the matrices $e_{ij}^m$.

Lemma 2. [2, Exercise from the 'Relative $K_1$' chapter] Let $n \ge 2$. Then we have $$\operatorname{SL}_n(\mathbb{Z}, 2\mathbb{Z}) = \operatorname{E}_n(\mathbb{Z}, 2\mathbb{Z}).$$

Lemma 3. Let $n \ge 2$. Then the quotient group $\overline{Q} := \operatorname{GL}_n(\mathbb{Z})/ \langle \langle \overline{\epsilon_1}, \dots, \overline{\epsilon_n} \rangle\rangle$ is isomorphic to $\operatorname{GL}_n(\mathbb{Z}/2\mathbb{Z})$.

Proof. Since both groups are generated by elementary matrices (i.e., the images of the automorphisms $\lambda_{ij}$) and the image of $\overline{\epsilon}_i$ is trivial in $\operatorname{GL}_n(\mathbb{Z}/2\mathbb{Z})$, the group $\overline{Q}$ naturally surjects onto $\operatorname{GL}_n(\mathbb{Z}/2\mathbb{Z})$. To show there is surjection from $\operatorname{GL}_n(\mathbb{Z}/2\mathbb{Z})$ onto $\overline{Q}$, write $$\operatorname{GL}_n(\mathbb{Z}) = \operatorname{SL}_n(\mathbb{Z}) \rtimes \langle \epsilon_1 \rangle$$ and observe that $[ \epsilon_i, e_{ij}] = e_{ij}^{2}$. We infer that $\langle\langle \overline{\epsilon}_1,\dots,\overline{\epsilon_n} \rangle\rangle$ contains $\operatorname{SL}_2(\mathbb{Z}, 2\mathbb{Z}) = \operatorname{E}_2(\mathbb{Z}, 2\mathbb{Z})$ (Lemma 2). Hence $\overline{Q}$ is a quotient of $\operatorname{SL}_n(\mathbb{Z})/\operatorname{E}_n(\mathbb{Z}, 2\mathbb{Z}) = \operatorname{GL}_n(\mathbb{Z}/2\mathbb{Z})$, which completes the proof.

We are now in position to prove the Claim.

Proof of the Claim. Clearly, the group $\overline{Q}$ of Lemma 1 is a quotient of $Q$. To see that $Q$ is a quotient of $\overline{Q}$ iff $(ii)$ holds true, combine Lemma 1 with Lemma 3 and the result of Nielsen asserting that $\operatorname{IA}(F_n) = \langle \langle \alpha_{112} \rangle \rangle$ for every $n \ge 2$.

Question 1. (HJRW) Can the Congruence Subroup Property of $\operatorname{SL}_n(\mathbb{Z})$ help prove the [3, Proposition 2.8] mentioned by the OP?

Answer. The ambient knowledge of the CSP encompasses Lemma 2, which certainly helps. That's however insufficient as the CSP has no bearing on Claim.$(ii)$.

Question 2. Can the Gersten Presentation of $\operatorname{Out}(F_n)$ [3, Proposition 2.5] help prove the statement $(ii)$ of the above Claim?

Answer. Yes. This is rather an afterthought inspired by Andy Putman's answer: the relations $\rho_{12}^{\epsilon_1} = \lambda_{12}^{-1}$ and $\rho_{21}^{\epsilon_1} = \rho_{21}^{-1}$ are indeed useful.

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• [1] R. Lyndon and P. Schupp, "Combinatorial Group Theory", 1977.
• [2], B. Magurn, An Algebraic introduction to K-theory", 2002.
• [3] D. Kielak, "Outer automorphism groups of free groups: linear and free representations", 2018.