Free products of free-group automorphism groups: a finite-rank embedding by separating cocycles

Problem

If $A,B\le\operatorname{Aut}(F_n)$ with $n\ge2$ , does $A*B$ embed in $\operatorname{Aut}(F_m)$ for some finite $m$ ?

A,B\le\operatorname{Aut}(F_n) \quad\stackrel{?}{\Longrightarrow}\quad A*B\hookrightarrow\operatorname{Aut}(F_m).

The construction below gives $m=2n+3$ and extends to finite indexed families with different initial ranks.

The embedding theorem

Write $F(X)$ for the free group on a set $X$ , and put $F_n=F(\{x_1,\ldots,x_n\})$ , with $F_0=1$ . Automorphisms act on the left, so $(\alpha\beta)(x)=\alpha(\beta(x))$ . All free products below are free products of groups.

Theorem 1. Let $I$ be a finite set, let $n_i\geq0$ for $i\in I$ , and let $G_i\leq\operatorname{Aut}(F_{n_i})$ . There is an injective homomorphism

\tag{1} \mathop{\ast}_{i\in I}G_i \longrightarrow \operatorname{Aut}(F_R), \qquad R=\sum_{i\in I}n_i+|I|+1.

The groups $G_i$ need not be finitely generated.

Corollary 2. If $G\leq\operatorname{Aut}(F_n)$ and $H\leq\operatorname{Aut}(F_m)$ , then

\tag{2} G*H\hookrightarrow\operatorname{Aut}(F_{n+m+3}).

In particular, if both groups are subgroups of $\operatorname{Aut}(F_n)$ , then $G*H\hookrightarrow\operatorname{Aut}(F_{2n+3})$ .

Proof. Apply Theorem 1 to a two-element index set. ◻

Problem 18.11, proposed by V. G. Bardakov, asks whether the free product of two subgroups of $\operatorname{Aut}(F_n)$ , for $n\geq2$ , embeds in $\operatorname{Aut}(F_m)$ for some finite $m$ (2, Problem 18.11). Corollary 2 supplies the explicit bound $m=2n+3$ . The construction also permits different initial ranks.

McCullough and Miller (4) study symmetric automorphisms of free products. A related embedding problem is treated by Marchand (3, Corollary 5.5), who constructs embeddings $\operatorname{Out}(Q)\hookrightarrow\operatorname{Out}(F)$ for suitable free products $Q$ of a free group and finite abelian groups, with $F$ a finite-index free subgroup of $Q$ . In Theorem 1, the source is a free product of subgroups of free-group automorphism groups, and the target is itself an automorphism group. We obtain the embedding from the cocycle criterion of Theorem 8 below.

Separating cocycles and free products

Definition 3. Let $\rho:A\to\operatorname{Aut}(B)$ be a homomorphism. A normalized nonabelian cocycle for $\rho$ is a map $v:A\to B$ satisfying

\tag{3} v(1)=1,\qquad v(gh)=v(g)\rho(g)(v(h))\quad(g,h\in A).

We call $v$ separating if $v(g)=1$ implies $g=1$ .

Let $I$ now be an arbitrary set. For each $i\in I$ , suppose that $A_i$ and $B_i$ are groups, $\alpha_i:A_i\to\operatorname{Aut}(B_i)$ is a homomorphism, and $c_i:A_i\to B_i$ satisfies

\begin{align*} c_i(1)&=1,\tag{4a}\\ c_i(ab)&=c_i(a)\alpha_i(a)(c_i(b)), &&a,b\in A_i,\tag{4b}\\ c_i(a)=1&\ \Longrightarrow\ a=1. \tag{4c} \end{align*}

Put

\tag{5} A=\mathop{\ast}_{i\in I}A_i, \qquad B=\mathop{\ast}_{i\in I}B_i, \qquad P=\prod_{i\in I}\operatorname{Aut}(B_i).

We identify the factors with their canonical images in the corresponding free products. For $p=(p_i)\in P$ , let $\overline p\in\operatorname{Aut}(B)$ be the automorphism whose restriction to $B_i$ is $p_i$ . The inverse is $\overline{p^{-1}}$ , and $\overline{pq}=\overline p\,\overline q$ . For $a\in A_i$ , define $\widehat\alpha_i(a)\in P$ by

\tag{6} \widehat\alpha_i(a)_j= \begin{cases} \alpha_i(a),&j=i,\\ \operatorname{id}_{B_j},&j\ne i. \end{cases}

Lemma 4. There are a homomorphism $\rho:A\to\operatorname{Aut}(B)$ and a normalized cocycle $v:A\to B$ such that

\tag{7} \rho(a)=\overline{\widehat\alpha_i(a)},\qquad v(a)=c_i(a)\quad(a\in A_i).

Every $\rho(g)$ preserves each free factor $B_i$ and restricts to an automorphism of that factor.

Proof. Use the componentwise action to form $B\rtimes P$ , with multiplication

\tag{8} (b,p)(d,q)=\bigl(b\,\overline p(d),pq\bigr).

Equations (4a) and (4b) give homomorphisms

A_i\longrightarrow B\rtimes P, \qquad a\longmapsto\bigl(c_i(a),\widehat\alpha_i(a)\bigr).

Their free-product extension is a homomorphism $L:A\to B\rtimes P$ . Write

L(g)=(v(g),\sigma(g)),\qquad \rho(g)=\overline{\sigma(g)}.

Projection onto $P$ shows that $\sigma$ , and hence $\rho$ , is a homomorphism. The first coordinate of $L(gh)=L(g)L(h)$ is exactly (3). The restrictions (7) follow from the definition of $L$ on each factor. Since $\sigma(g)\in P$ , the last assertion follows from the definition of the componentwise action. ◻

Lemma 5. The cocycle $v$ in Lemma 4 is separating.

Proof. Let $g\ne1$ and write its reduced free-product normal form as

g=a_1\cdots a_\ell, \qquad \ell\geq1,\quad a_r\in A_{i_r}\setminus\{1\},\quad i_r\ne i_{r+1}\quad(1\leq r<\ell).

Set $g_0=1$ and $g_r=a_1\cdots a_r$ . Repeated application of (3) gives

\tag{12} v(g)=b_1\cdots b_\ell, \qquad b_r=\rho(g_{r-1})\bigl(c_{i_r}(a_r)\bigr).

By (4c), $c_{i_r}(a_r)\ne1$ . Lemma 4 therefore gives $b_r\in B_{i_r}\setminus\{1\}$ . The indices $i_1,\ldots,i_\ell$ are unchanged, so the right-hand side of (12) is a nonempty reduced word in $B$ . The free-product normal-form theorem implies $v(g)\ne1$ . ◻

One additional free generator

Let $C$ be a group and put $C^+=C*\langle z\rangle$ , where $\langle z\rangle$ is infinite cyclic.

The following added-generator construction is standard. Automorphisms fixing $C$ and sending $z$ to $zu$ occur, for example, in Bardakov and Mikhailov (1, proof of Theorem 4).

Lemma 6. For $\alpha\in\operatorname{Aut}(C)$ and $u\in C$ , the assignments

\tag{13} T_{\alpha,u}(c)=\alpha(c)\quad(c\in C), \qquad T_{\alpha,u}(z)=zu

define an automorphism of $C^+$ . These automorphisms satisfy

\begin{align*} T_{\alpha,u}T_{\beta,w} &=T_{\alpha\beta,\,u\alpha(w)},\tag{14a}\\ T_{\alpha,u}^{-1} &=T_{\alpha^{-1},\,\alpha^{-1}(u^{-1})}. \tag{14b} \end{align*}

Proof. The universal property of $C*\langle z\rangle$ first gives a homomorphism $T_{\alpha,u}$ with the stated values. On $C$ , the composite in (14a) restricts to $\alpha\beta$ . On the remaining generator,

(T_{\alpha,u}T_{\beta,w})(z) =T_{\alpha,u}(zw)=zu\alpha(w).

The same universal property proves (14a). Moreover, $T_{\operatorname{id},1}=\operatorname{id}_{C^+}$ , and the two choices in (14b) give, in the two orders, the pairs

\begin{aligned} \bigl(\operatorname{id},u\alpha(\alpha^{-1}(u^{-1}))\bigr)&=(\operatorname{id},1),\\ \bigl(\operatorname{id},\alpha^{-1}(u^{-1})\alpha^{-1}(u)\bigr)&=(\operatorname{id},1). \end{aligned}

Thus (14b) is a two-sided inverse. ◻

Proposition 7. Let $\rho:A\to\operatorname{Aut}(C)$ be a homomorphism and let $v:A\to C$ be a normalized cocycle. Then

\tag{17} \Theta:A\longrightarrow\operatorname{Aut}(C*\langle z\rangle), \qquad \Theta(g)=T_{\rho(g),v(g)}

is a homomorphism. If $v$ is separating, then $\Theta$ is injective.

Proof. The identity element is preserved because $\rho(1)=\operatorname{id}$ and $v(1)=1$ . By (14a) and the cocycle identity,

\Theta(g)\Theta(h) =T_{\rho(g)\rho(h),\,v(g)\rho(g)(v(h))} =T_{\rho(gh),v(gh)}=\Theta(gh).

If $\Theta(g)=\operatorname{id}$ , evaluation at $z$ gives $zv(g)=z$ . Cancellation, followed by injectivity of the canonical map $C\to C*\langle z\rangle$ , gives $v(g)=1$ . If $v$ is separating, then $g=1$ , proving injectivity. ◻

Theorem 8. Let $I$ be any set, and suppose that the groups $A_i,B_i$ , homomorphisms $\alpha_i:A_i\to\operatorname{Aut}(B_i)$ , and maps $c_i:A_i\to B_i$ satisfy (4a)–(4c). There is an injective homomorphism

\tag{19} \mathop{\ast}_{i\in I}A_i \longrightarrow \operatorname{Aut}\!\left(\left(\mathop{\ast}_{i\in I}B_i\right) *\langle z\rangle\right).

Its restriction to $A_i$ sends $a$ to the automorphism whose restriction to $B_i$ is $\alpha_i(a)$ , whose restriction to every $B_j$ with $j\ne i$ is the identity, and which sends $z$ to $zc_i(a)$ .

Proof. Lemmas 4 and 5 give an action $\rho$ on $B=\mathop{\ast}_{i\in I}B_i$ and a separating cocycle $v:A\to B$ . Apply Proposition 7. Equation (7) gives the stated restrictions. ◻

Markers for free-group automorphisms

The separating condition in Theorem 8 can be obtained from a point with trivial stabilizer.

Lemma 9. Let $\alpha:A\to\operatorname{Aut}(C)$ be a homomorphism and let $w\in C$ . Then

\tag{20} c(a)=w^{-1}\alpha(a)(w)

is a normalized cocycle. It is separating if the stabilizer of $w$ under $\alpha$ is trivial.

Proof. We have $c(1)=w^{-1}w=1$ , and

\begin{align*} c(a)\alpha(a)(c(b)) &=w^{-1}\alpha(a)(w)\, \alpha(a)\bigl(w^{-1}\alpha(b)(w)\bigr)\\ &=w^{-1}\alpha(ab)(w)=c(ab). \end{align*}

Also $c(a)=1$ is equivalent to $\alpha(a)(w)=w$ , which gives the last assertion. ◻

For a group $C$ and a finite list $(y_1,\ldots,y_r)$ of its elements, define

\tag{22} \mu_C(y_1,\ldots,y_r)=y_1t\cdots y_rt \in C*\langle t\rangle, \qquad \mu_C(\varnothing)=1,

where $\langle t\rangle$ is infinite cyclic.

Lemma 10. If every $y_j$ and every $y'_j$ is nonidentity, then

\mu_C(y_1,\ldots,y_r)=\mu_C(y'_1,\ldots,y'_s) \quad\Longrightarrow\quad r=s\ \text{ and }\ y_j=y'_j\ (1\leq j\leq r).

Proof. For $r>0$ , the expression $y_1t\cdots y_rt$ is a reduced free-product word of syllable length $2r$ : its syllables are nonidentity and alternate between $C$ and $\langle t\rangle$ . The same holds on the right when $s>0$ . Uniqueness of reduced normal forms gives equal lengths and equal corresponding syllables. If either list is empty, its product is the identity; the other list must then be empty as well. ◻

Lemma 11. Let $X=\{x_1,\ldots,x_n\}$ be a free basis, put $B=F(X)*\langle t\rangle$ , and set

\tag{24} w=x_1t\cdots x_nt.

Extend each $\alpha\in\operatorname{Aut}(F(X))$ to $\overline\alpha\in\operatorname{Aut}(B)$ by fixing $t$ . If $\overline\alpha(w)=w$ , then $\alpha=\operatorname{id}$ .

Proof. The free-product universal property gives the extension and the identity $\overline{\alpha\beta}=\overline\alpha\,\overline\beta$ . Each $x_j$ is nonidentity, so each $\alpha(x_j)$ is nonidentity. The assumed equality reads

\alpha(x_1)t\cdots\alpha(x_n)t=x_1t\cdots x_nt.

Lemma 10 gives $\alpha(x_j)=x_j$ for every $j$ . An endomorphism of a free group is determined by its values on the free basis, hence $\alpha=\operatorname{id}$ . The same argument applies when $n=0$ : the basis is empty and the free group has only its identity automorphism. ◻

The finite-family construction and its rank

Proof of Theorem 1. Choose pairwise disjoint bases $X_i=\{x_{i,1},\ldots,x_{i,n_i}\}$ , marker letters $t_i$ , and one further letter $z$ . Define

\tag{26} B_i=F(X_i)*\langle t_i\rangle, \qquad w_i=x_{i,1}t_i\cdots x_{i,n_i}t_i.

Extend $g\in G_i$ to $\overline g\in\operatorname{Aut}(B_i)$ by fixing $t_i$ , and put $c_i(g)=w_i^{-1}\overline g(w_i)$ . By Lemmas 9 and 11,

\tag{27} c_i(1)=1,\qquad c_i(gh)=c_i(g)\overline g(c_i(h)),\qquad c_i(g)=1\Longrightarrow g=1.

Theorem 8 therefore gives an embedding into $\operatorname{Aut}(E)$ , where

E=\left(\mathop{\ast}_{i\in I}B_i\right)*\langle z\rangle.

The automorphism assigned to $g\in G_i$ is determined by

\begin{align*} \Theta_i(g)|_{B_i}&=\overline g, &\Theta_i(g)|_{B_j}&=\operatorname{id}_{B_j}\quad(j\ne i), \tag{29a}\\ \Theta_i(g)(z)&=z\,w_i^{-1}\overline g(w_i). \tag{29b} \end{align*}

It remains to identify $E$ . Let

Y=\left(\coprod_{i\in I}X_i\right) \amalg\{t_i:i\in I\}\amalg\{z\}.

The inclusions of the named generators give a homomorphism $F(Y)\to E$ . Conversely, their images in $F(Y)$ give homomorphisms from every $B_i$ and from $\langle z\rangle$ , hence a homomorphism $E\to F(Y)$ . The two composites fix every named generator, so both are identity homomorphisms by the respective universal properties. Thus

\tag{31} E\cong F(Y),\qquad |Y|=\sum_{i\in I}(n_i+1)+1 =\sum_{i\in I}n_i+|I|+1=R.

Choose an isomorphism $e:E\to F_R$ . The map $\gamma\mapsto e\gamma e^{-1}$ is an isomorphism $\operatorname{Aut}(E)\to\operatorname{Aut}(F_R)$ ; composing it with the constructed embedding proves (1). ◻

The rank calculation applies to an empty family as well: the source is the trivial group and $E=\langle z\rangle$ . Zero ranks cause no change to the proof, since $\operatorname{Aut}(F_0)=1$ and the corresponding marker is the empty product.

For comparison, let $k=|I|\geq2$ . Iterate Corollary 2, identifying each preceding free product with its embedded image before adjoining the next factor. The iterated and simultaneous bounds are

\begin{align*} R_{\mathrm{bin}}&=\sum_{i\in I}n_i+3(k-1), \tag{32a}\\ R&=\sum_{i\in I}n_i+k+1, \qquad R_{\mathrm{bin}}-R=2k-4. \tag{32b} \end{align*}

Thus finite-family existence already follows by binary iteration; using one common generator $z$ saves $2k-4$ generators in this comparison. No minimality is asserted. For a singleton family the given inclusion $G_1\leq\operatorname{Aut}(F_{n_1})$ already uses fewer generators.

References

Preprint · Lean (GitHub)

V. G. Bardakov and R. Mikhailov, On certain questions of the free group automorphisms theory, Comm. Algebra 36 (2008), no. 4, 1489–1499. doi:10.1080/00927870701866929.

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st ed., version 46 (2026), Problem 18.11. arXiv:1401.0300v46.

A. Marchand, Free representations of outer automorphism groups of free products via characteristic abelian coverings, J. Group Theory 26 (2023), no. 2, 399–420. doi:10.1515/jgth-2021-0154.

D. McCullough and A. Miller, Symmetric automorphisms of free products, Mem. Amer. Math. Soc. 122 (1996), no. 582, viii+97 pp. doi:10.1090/memo/0582.