Automorphisms of pure braid Groups

TL;DR

This paper characterizes the automorphism groups of pure braid groups for n>3, showing they are generated by specific subgroups and an extra automorphism, and explores automorphism extension problems with negative results.

Contribution

It provides a detailed description of automorphism groups of pure braid groups and addresses the extension and lifting problems for these automorphisms.

Findings

Automorphism group of P_n for n>3 is generated by specific subgroups and an extra automorphism.

No non-trivial central automorphism of P_n extends to B_n.

Extension and lifting problems for automorphisms mostly have negative solutions.

Abstract

In this paper, we investigate the structure of the automorphism groups of pure braid groups. We prove that, for $n>3$, $\Aut(P_n)$ is generated by the subgroup $\Aut_c(P_n)$ of central automorphisms of $P_n$, the subgroup $\Aut(B_n)$ of restrictions of automorphisms of $B_n$ on $P_n$ and one extra automorphism $w_n$. We also investigate the lifting and extension problem for automorphisms of some well-known exact sequences arising from braid groups, and prove that that answers are negative in most cases. Specifically, we prove that no non-trivial central automorphism of $P_n$ can be extended to an automorphism of $B_n$.

Equations

$$\mathrm{Aut}(B_n)\cong \mathrm{Inn}(B_n)⋊{\mathbb{Z}}_2,$$

$$\mathrm{Aut}(P_n)\cong {\mathrm{Aut}}_c(P_n)⋊\mathrm{Aut}({\overline{P}}_n),$$

$$\mathrm{Aut}(B_n)\cong \mathrm{Mod}({\mathbb{D}}_n),$$

$$\{{\sigma }_1,{\sigma }_2,\dots ,{\sigma }_{n-1}\},$$

$${\sigma }_i{\sigma }_j$$

$${\sigma }_i{\sigma }_{i+1}{\sigma }_i$$

$$A_{i,j}={\sigma }_{j-1}{\sigma }_{j-2}\dots {\sigma }_{i+1}{\sigma }_i^2{\sigma }_{i+1}^{-1}\dots {\sigma }_{j-2}^{-1}{\sigma }_{j-1}^{-1}\text{where}1\le i<j\le n,$$

$$A_{i,k}{A}_{ij}{A}_{k,j}$$

$$A_{l,j}{A}_{k,l}{A}_{k,j}$$

$$(A_{k,l}{A}_{k,j}{A}_{k,l}^{-1})A_{i,l}$$

$$A_{k,j}{A}_{i,l}$$

$${\sigma }_k^{-\epsilon }{A}_{i,j}{\sigma }_k^{\epsilon }$$

$${\sigma }_i^{-\epsilon }{A}_{i,i+1}{\sigma }_i^{\epsilon }$$

$${\sigma }_{i-1}^{-1}{A}_{i,j}{\sigma }_{i-1}$$

$${\sigma }_{i-1}{A}_{i,j}{\sigma }_{i-1}^{-1}$$

$${\sigma }_i^{-1}{A}_{i,j}{\sigma }_i$$

$${\sigma }_i{A}_{i,j}{\sigma }_i^{-1}$$

$${\sigma }_{j-1}^{-1}{A}_{i,j}{\sigma }_{j-1}$$

$${\sigma }_{j-1}{A}_{i,j}{\sigma }_{j-1}^{-1}$$

$${\sigma }_j^{-1}{A}_{i,j}{\sigma }_j$$

$${\sigma }_j{A}_{i,j}{\sigma }_j^{-1}$$

$$U_i=⟨A_{1,i},A_{2,i},\dots ,A_{i-1,i}⟩,$$

$$A_{i,k}^{-\epsilon }{A}_{k,j}{A}_{i,k}^{\epsilon }$$

$$A_{k,l}^{-\epsilon }{A}_{k,j}{A}_{k,l}^{\epsilon }$$

$$A_{i,l}^{-\epsilon }{A}_{k,j}{A}_{i,l}^{\epsilon }$$

$$1\to U_n\to P_n\to P_{n-1}\to 1.$$

$$P_n\cong U_n⋊(U_{n-1}⋊(\cdots ⋊(U_3⋊U_2)\cdots )).$$

$$z_n=A_{1,2}(A_{1,3}{A}_{2,3})\cdots (A_{1,n}\cdots A_{n-1,n}).$$

$$P_n\cong Z(P_n)\times ⟨A_{1,3},A_{2,3},\dots ,A_{1,n},\dots ,A_{n-1,n}⟩,$$

$$A_{i,j}\mapsto A_{i,j}{z}_n^{t_{ij}},$$

$${\mathrm{Aut}}_c(P_n)\to {\mathbb{Z}}_2$$

$${\mathrm{Aut}}_c(P_n)\cong {\mathbb{Z}}^N⋊{\mathbb{Z}}_2.$$

$$\psi:A_{p,q}\mapsto\left\{\begin{array}[]{ll}A_{1,2}z_{n}^{-2}&~{}\textrm{for}~{}p=1,q=2\\ A_{p,q}&~{}\textrm{otherwise}\\ \end{array}\right.$$

$$\phi_{i,j}:A_{p,q}\mapsto\left\{\begin{array}[]{ll}A_{1,2}z_{n}&~{}\textrm{for}~{}p=1,q=2\\ A_{i,j}z_{n}^{-1}&~{}\textrm{for}~{}p=i,q=j\\ A_{p,q}&~{}\textrm{otherwise}.\\ \end{array}\right.$$

$$\begin{array}[]{ll}\omega_{i}\omega_{j}=\omega_{j}\omega_{i}\ \text{for}\ |i-j|>1,\\ \omega_{i}\omega_{i+1}\omega_{i}=\omega_{i+1}\omega_{i}\omega_{i+1},\\ \omega_{1}\cdots\omega_{n-1}\omega_{n}^{2}\omega_{n-1}\cdots\omega_{1}=1,\\ (\omega_{1}\omega_{2}\cdots\omega_{n})^{n+1}=1,\\ (\varepsilon\omega_{i})^{2}=1,~{}\textrm{and}~{}\varepsilon^{2}=1.\end{array}$$

$$\omega_{k}\colon A_{i,j}\mapsto\left\{\begin{array}[]{ll}A_{i-1,j}&\text{if $k=i-1$}\\ A_{i+1,j}^{A_{i,i+1}}&\text{if $k=i<j-1$}\\ A_{i,j-1}&\text{if $k=j-1>i$}\\ A_{i,j+1}^{A_{j,j+1}}&\text{if $k=j$}\\ A_{i,j}&\text{otherwise},\end{array}\right.$$

$$\omega_{2}\colon A_{i,j}\mapsto\left\{\begin{array}[]{ll}A_{1,3}^{A_{2,3}}z_{n}&\text{if $i=1$, $j=2$}\\ A_{1,2}z_{n}^{-1}&\text{if $i=1$, $j=3$}\\ A_{3,j}^{A_{2,3}}&\text{if $i=2$, $j\geq 4$}\\ A_{2,j}&\text{if $i=3$}\\ A_{i,j}&\text{otherwise,}\end{array}\right.$$

Full text

1. Introduction

Let $B_{n}$ be the Artin braid group on $n$-strings, and $\{\sigma_{1},\dots,\sigma_{n-1}\}$ the set of standard generators of $B_{n}$. In [1], Artin mentioned the problem of determining all automorphisms of the braid groups. Subsequently, in [2], he determined all representations of $B_{n}$ by transitive permutation groups in $n$ letters. Since then the subject has attracted a lot of attention. In [6], Dyer and Grossman completely determined the automorphism groups of braid groups. More precisely, they proved that

[TABLE]

where $\operatorname{Inn}(B_{n})\cong B_{n}/Z(B_{n})$ is the group of inner automorphisms and $\mathbb{Z}_{2}=\langle\tau\rangle$ with $\tau(\sigma_{i})=\sigma_{i}^{-1}$ for all $1\leq i\leq n-1$.

The pure braid group $P_{n}$ is the kernel of the natural homomorphism $B_{n}\to S_{n}$, from the braid group to the symmetric group. Once the structure of $\operatorname{Aut}(B_{n})$ is revealed, it is natural to investigate the structure of $\operatorname{Aut}(P_{n})$. It is well known that $P_{n}\cong Z(P_{n})\times\overline{P}_{n}$, where $\overline{P}_{n}=P_{n}/Z(P_{n})$. Using this decomposition, Bell and Margalit [3, Theorem 8] proved that, for $n\geq 4$,

[TABLE]

where $\operatorname{Aut}(\overline{P}_{n})\cong\operatorname{Mod}(\mathbb{S}_{n+1})$. Here $\operatorname{Aut}_{c}(P_{n})$ is the group of central automorphisms of $P_{n}$ and $\operatorname{Mod}(\mathbb{S}_{n+1})$ is the extended mapping class group of the 2-sphere with $n+1$ punctures. Hence $\operatorname{Aut}(P_{n})\cong\operatorname{Aut}_{c}(P_{n})\rtimes\operatorname{Mod}(\mathbb{S}_{n+1})$ for $n\geq 4$. This result was used by Cohen [5] to give an explicit presentation of the automorphism group $\operatorname{Aut}(P_{n})$ for $n\geq 3$. Notice that from this point of view the result of Dyer and Grossman has the form

[TABLE]

where $\mathbb{D}_{n}$ is the disc with $n$ punctures.

Since the group of pure braids $P_{n}$ is the direct product of the center $Z(P_{n})$ and the group $\overline{P}_{n}$, it is not difficult to determine the group of central automorphisms $\operatorname{Aut}_{c}(P_{n})$. On the other hand, $P_{n}$ is a characteristic subgroup of $B_{n}$, and hence each automorphism of $B_{n}$ induces an automorphism of $P_{n}$. This leads to the following question:

Question 1**.**

Is $\operatorname{Aut}(P_{n})$ generated by $\operatorname{Aut}_{c}(P_{n})$ and $\operatorname{Aut}(B_{n})$?

After recalling some preliminary results in Section 2, we answer the preceding question in Sections 3 and 4 of the paper. The answer is evidently affirmative for $n=2$, and we prove that it is so for $n=3$ as well (Proposition 4.7). For $n>3$, we prove that $\operatorname{Aut}(P_{n})$ is generated by $\operatorname{Aut}_{c}(P_{n})$, $\operatorname{Aut}(B_{n})$ and one extra automorphism $w_{n}$ (Theorem 3.6).

The last two sections of the paper are dedicated to the lifting and extension of automorphisms in certain short exact sequences arising from braid groups. A general formulation is the following:

Question 2**.**

Let $1\to K\to G\to H\to 1$ be a short exact sequence of groups. Given $\phi\in\operatorname{Aut}(H)$, does there exists an automorphism of $G$ which induces $\phi?$ Analogously, given $\psi\in\operatorname{Aut}(K)$, does there exists an automorphism of $G$ whose restriction to $K$ is $\psi?$

For finite groups, this problem has been investigated in [7, 10, 11] using cohomological methods and the fundamental exact sequence of Wells [12]. In Section 5, we prove that the answers to both the questions are negative (Theorems 5.1 and 5.6) for the extension $1\to U_{4}\to P_{4}\to P_{3}\to 1$. Finally, in Section 6, we consider the extension $1\to P_{n}\to B_{n}\to S_{n}\to 1$. We prove that the non-inner automorphism of $S_{6}$ cannot be lifted to an automorphism of $B_{6}$ (Proposition 6.2), and that no non-trivial element of $\operatorname{Aut}_{c}(P_{n})$ can be extended to an automorphism of $B_{n}$ (Theorem 6.6).

2. Preliminary results

We use the standard notations. Given two elements $x,y$ of a group $G$, we write $y^{x}=x^{-1}yx$ and $[x,y]=x^{-1}y^{-1}xy$. Further, for $z\in G$, $\hat{z}$ denotes the inner automorphism of $G$ induced by $z$. An automorphism of a group $G$ is called central if it induces identity automorphism on the central quotient. The subgroup of all central automorphisms of $G$ is denoted by $\operatorname{Aut}_{c}(G)$.

Next we recall some basic facts on braid groups and refer the reader to [4, 9] for more details. The braid group $B_{n}$, $n\geq 2$, on $n$-strings is generated by the set

[TABLE]

and is defined by relations

[TABLE]

Note that $B_{1}$ is defined as the trivial group and $B_{2}$ turns out to be the infinite cyclic group. The subgroup of $B_{n}$ generated by the elements

[TABLE]

is called the pure braid group, and denoted by $P_{n}$. Further, $P_{n}$ is defined by the relations

[TABLE]

It is readily seen that $P_{1}$ is the trivial group, $P_{2}\cong\mathbb{Z}$, and $P_{3}\cong F_{2}\times\mathbb{Z}$. In general, $P_{n}$ is characteristic in $B_{n}$, and the quotient $B_{n}/P_{n}$ is the symmetric group $S_{n}$. Further, the generators of $B_{n}$ act on the generators $A_{i,j}\in P_{n}$ by the following rules:

[TABLE]

where $\varepsilon=\pm 1$.

For each $i=2,\ldots,n$, consider the following subgroup

[TABLE]

of $P_{n}$. It is known that each $U_{i}$ is a free subgroup of rank $i-1$. One can rewrite the relations of $P_{n}$ as the following conjugation rules (for $\varepsilon=\pm 1$):

[TABLE]

It follows from these rules that $U_{n}$ is normal in $P_{n}$, and $P_{n}$ has the decomposition $P_{n}=U_{n}\rtimes P_{n-1}$. This gives rise to the split extension

[TABLE]

By induction on $n$, it follows that $P_{n}$ is the semi-direct products of free groups as follows

[TABLE]

It is known that, for $n\geq 2$, the center $Z(P_{n})=\langle z_{n}\rangle$ is infinite cyclic generated by the full twist braid

[TABLE]

It follows that $P_{n}\cong Z(P_{n})\times\overline{P}_{n}$, where $\overline{P}_{n}=P_{n}/Z(P_{n})$ (see [4]). More precisely,

[TABLE]

Using this decomposition, Bell and Margalit [3, Theorem 8] proved the following result.

Theorem 2.1**.**

$\operatorname{Aut}(P_{n})\cong\operatorname{Aut}_{c}(P_{n})\rtimes\operatorname{Aut}(\overline{P}_{n})$* for $n\geq 4$.*

Next, we recall a presentation for $\operatorname{Aut}(P_{n})$ from Cohen [5]. The subgroup of central automorphisms $\operatorname{Aut}_{c}(P_{n})$ consists of automorphisms of the form

[TABLE]

where $t_{ij}\in\mathbb{Z}$ and $\sum t_{ij}=0$ or -2. We have $z_{n}\mapsto z_{n}$ in the former case, and $z_{n}\mapsto z_{n}^{-1}$ in the latter case. This gives a surjection

[TABLE]

with kernel consisting of automorphisms for which $\sum t_{ij}=0$. Since $P_{n}$ has $\binom{n}{2}$ generators, this kernel is free abelian of rank $N$, where $N=\binom{n}{2}-1$. Further, the choice $t_{12}=-2$ and all other $t_{ij}=0$ gives a splitting $\mathbb{Z}_{2}\to\operatorname{Aut}_{c}(P_{n})$, and hence

[TABLE]

As in [5, Equation (11)], we note that this group is generated by the automorphisms $\psi,\phi_{i,j}:P_{n}\to P_{n}$, $1\leq i<j\leq n$, $\{i,j\}\neq\{1,2\}$, where

[TABLE]

and

[TABLE]

It can be easily checked that $\psi^{2}=1$ and $\psi\phi_{i,j}\psi=\phi_{i,j}^{-1}$.

Let $\mathbb{S}_{n+1}$ denote the 2-sphere with $n+1$ punctures, and $\operatorname{Mod}(\mathbb{S}_{n+1})$ its extended mapping class group. For $n\geq 1$, $\operatorname{Mod}(\mathbb{S}_{n+1})$ has a presentation with generators $\{\omega_{1},\dots,\omega_{n},\varepsilon\}$ and relations

[TABLE]

We refer the reader to [4, Theorem 4.5] for details. By Bell and Margalit [3], $\operatorname{Mod}(\mathbb{S}_{n+1})\cong\operatorname{Aut}(\overline{P}_{n})$ and can be viewed as a subgroup of $\operatorname{Aut}(P_{n})$. Further, the generators $\omega_{1},\omega_{2},\dots,\omega_{n},\varepsilon$ act as automorphisms of $P_{n}$ in the following manner [5, Equation (12)]:

[TABLE]

for  $1\leq k\leq n-1$  and $k\neq 2$;

[TABLE]

[TABLE]

and

[TABLE]

3. Automorphism group of $P_{n}$ for $n\geq 4$

As we noted in the introduction $\operatorname{Aut}(B_{n})=\langle\operatorname{Inn}(B_{n}),\tau\rangle$, and each automorphism of $B_{n}$ induces an automorphism of $P_{n}$. Let $t=\tau|_{P_{n}}$ and $s_{i}=\hat{\sigma_{i}}|_{P_{n}}$ for all $1\leq i\leq n-1$, where $\hat{\sigma_{i}}$ is the inner automorphism of $B_{n}$ induced by $\sigma_{i}$.

Lemma 3.1**.**

The automorphism $\tau$ induces the automorphism $t$ of $P_{n}$ which acts on the generators by the rule

[TABLE]

for all $1\leq i<j\leq n$.

Proof.

We use induction on $n$. Notice that it suffices to prove the equality

[TABLE]

For $n=3$ we have

[TABLE]

Suppose that the formula holds for $n$. Then

[TABLE]

This proves the lemma. ∎

An immediate consequence is the following.

Corollary 3.2**.**

The following equalities hold

[TABLE]

i.e. $t\varepsilon=\psi.$

It is not difficult to see that the restriction homomorphism $\operatorname{Aut}(B_{n})\to\operatorname{Aut}(P_{n})$ is an embedding, enabling us to view $\operatorname{Aut}(B_{n})$ as a subgroup of $\operatorname{Aut}(P_{n})$. Now, to determine $\operatorname{Aut}(P_{n})$, introduce the following automorphism of $P_{n}$:

[TABLE]

Since the maps $\omega_{n}$ and $w_{n}$ differ by a central automorphism, it follows that $w_{n}$ also is an automorphism of the group $P_{n}$.

Lemma 3.3**.**

The automorphism $w_{n}$ does not lie in the subgroup $\langle\operatorname{Aut}_{c}(P_{n}),\operatorname{Aut}(B_{n})\rangle$ for $n\geq 4$.

Proof.

The automorphism $w_{n}$ induces an automorphism $\overline{w}_{n}$ of the abelianisation $P_{n}/P_{n}^{\prime}$. We prove that $\overline{w}_{n}$ does not lies in the image of $\langle\operatorname{Aut}_{c}(P_{n}),\operatorname{Aut}(B_{n})\rangle$ into $\operatorname{Aut}(P_{n}/P_{n}^{\prime})$. Indeed, since $\operatorname{Aut}(B_{n})=\left\langle\operatorname{Inn}(B_{n}),\,\,\tau\right\rangle$, the actions of elements of $\operatorname{Aut}(B_{n})$ on the quotient $P_{n}/P_{n}^{\prime}$ are compositions of inversion of $A_{ij}$ and permutations of these elements. Hence, it is enough to show that, in the quotient $P_{n}/P_{n}^{\prime}$, the set consisting of

[TABLE]

[TABLE]

[TABLE]

[TABLE]

cannot be obtained from the set

[TABLE]

using composition of the maps

[TABLE]

where $\alpha_{ij}=\pm 1$, $\beta_{ij}\in\mathbb{Z}$, $\left\{i,j\right\}\mapsto\left\{k,l\right\}$ is the permutation of pairs of indexes $\left\{i,j\right\}$, $1\leq i<j\leq n$. Note that in this case $z_{n}\mapsto z_{n}^{\pm 1}$. Let us take

[TABLE]

If the transformation $w_{n}$ modulo $P_{n}^{\prime}$ is a composition of maps of the specified type, then the word

[TABLE]

with a suitable $\beta\in\mathbb{Z}$ has the form $A_{k,l}^{\pm 1}$, but it is not true. Hence $w_{n}$ does not lie in $\langle\operatorname{Aut}_{c}(P_{n}),\operatorname{Aut}(B_{n})\rangle$. ∎

Corollary 3.4**.**

The subgroup $\left\langle\operatorname{Aut}_{c}(P_{n}),\,\,\operatorname{Aut}(B_{n})\right\rangle$ is not normal in the group $\operatorname{Aut}(P_{n})$ for $n\geq 4$.

Proof.

Set $H_{n}=\left\langle\operatorname{Aut}_{c}(P_{n}),\,\,\operatorname{Aut}(B_{n})\right\rangle$, and suppose that $H_{n}$ is normal in $\operatorname{Aut}(P_{n})$. Then the automorphism $w_{n}$ induces an automorphism of $\operatorname{Aut}(B_{n})\cong H_{n}/\operatorname{Aut}_{c}(P_{n})$. By [6, Theorem 22], the group $\operatorname{Aut}(B_{n})$ is complete for $n\geq 4$, i.e. has trivial center and only inner automorphisms. Since $\operatorname{Aut}(B_{n})=\left\langle\operatorname{Inn}(B_{n}),\,\,\tau\right\rangle$, $\operatorname{Aut}(B_{n})/\operatorname{Inn}(B_{n})\cong\mathbb{Z}_{2}$ and $w_{n}\not\in H_{n}$ by Lemma 3.3, it follows that the product $w_{n}t$ must lies in the subgroup generated by inner automorphism group $\operatorname{Inn}(B_{n})$ and central automorphism group $\operatorname{Aut}_{c}(P_{n})$, which is a contradiction. Hence $H_{n}$ is not normal in $\operatorname{Aut}(P_{n})$ for $n\geq 4$. ∎

Let $G_{n}$ be the subgroup of $\operatorname{Aut}(P_{n})$ consisting of those automorphisms which are restrictions of automorphisms of $B_{n}$ and the automorphism $w_{n}$, i.e.

[TABLE]

Lemma 3.5**.**

The automorphisms $\omega_{1},\omega_{2},\dots,\omega_{n},\varepsilon$ lie in $\langle\operatorname{Aut}_{c}(P_{n}),G_{n}\rangle$.

Proof.

It follows that $\omega_{k}(z_{n})=z_{n}$ for all $1\leq k\leq n$, and $\varepsilon(z_{n})=z_{n}$. As noted by Cohen [5], for $1\leq k\leq n-1$ and $k\neq 2$, the automorphism $\omega_{k}$ is given by the conjugation action of the braid $\sigma_{k}$ on the pure braid group. More precisely, $\omega_{k}(A_{i,j})=A_{i,j}^{\sigma_{k}}=\sigma_{k}^{-1}A_{i,j}\sigma_{k}$. On the other hand, $\omega_{2}$ is the composite of the conjugation action of $\sigma_{2}$ and the automorphism $\phi_{1,3}$ given by

[TABLE]

The automorphism $\omega_{n}$ is the product of $w_{n}$ and some central automorphism. Finally, the automorphism $\varepsilon$ is the product of $t$ and some central automorphism. ∎

We now prove the main result of this section.

Theorem 3.6**.**

$\operatorname{Aut}(P_{n})\cong\operatorname{Aut}_{c}(P_{n})\rtimes G_{n}$* for $n\geq 4$.*

Proof.

Obviously, $\operatorname{Aut}_{c}(P_{n})\rtimes G_{n}\leq\operatorname{Aut}(P_{n})$. By Bell and Margalit [3, Theorem 8], $\operatorname{Aut}(P_{n})\cong\operatorname{Aut}_{c}(P_{n})\rtimes\operatorname{Mod}(\mathbb{S}_{n+1})$ for $n\geq 4$. Recall that, $\operatorname{Mod}(\mathbb{S}_{n+1})=\langle\omega_{1},\dots,\omega_{n},\varepsilon\rangle$. The proof of the theorem is now complete by Lemma 3.5. ∎

We conclude this section with an explicit presentation of $\operatorname{Aut}(P_{4})$. By definition of $\omega_{i}$’s and $\varepsilon$, we have

[TABLE]

Now rewriting the relations of $\operatorname{Aut}(P_{4})$ in the new generators yields the following result.

Proposition 3.7**.**

The group $\operatorname{Aut}(P_{4})$ in the generators

[TABLE]

is defined by the relations:

[TABLE]

[TABLE]

4. Automorphism group of $P_{3}$

For $n=3$, we have $B_{3}=\langle\sigma_{1},\sigma_{2}~{}|~{}\sigma_{2}\sigma_{1}\sigma_{2}=\sigma_{1}\sigma_{2}\sigma_{1}\rangle$. Let $A_{1,2}=\sigma_{1}^{2}$, $A_{1,3}=\sigma_{2}\sigma_{1}^{2}\sigma_{2}^{-1}$ and $A_{2,3}=\sigma_{2}^{2}$. Then

[TABLE]

We set $x=A_{1,3}$, $y=A_{2,3}$ and $z=A_{1,2}A_{1,3}A_{2,3}$. Then $P_{3}=\langle z\rangle\times\langle x,y\rangle$, where $\langle x,y\rangle$ is the free group $F_{2}$ on two generators.

Lemma 4.1**.**

The action of $\sigma_{1}$ and $\sigma_{2}$ on $x,y,z$ is given by

1. (1)

$z^{\sigma_{1}}=z^{\sigma_{2}}=z$. 2. (2)

$x^{\sigma_{1}}=xyx^{-1}$* and $x^{\sigma_{2}}=y^{-1}x^{-1}z$.* 3. (3)

$y^{\sigma_{1}}=x$* and $y^{\sigma_{2}}=y$.*

Proof.

(1) follows trivially since $z\in Z(B_{3})$. For (2), we consider

[TABLE]

The rest of the identities follow easily. ∎

Remark 4.2**.**

The preceding lemma also implies that the automorphisms of $P_{3}$ induced by $\sigma_{1}$ and $\sigma_{2}$ are, in fact, automorphisms of $F_{2}=\langle x,y\rangle$.

Next, we describe $\operatorname{Aut}(P_{3})$. Consider the following automorphisms of $P_{3}$

[TABLE]

[TABLE]

where $\psi$ is an arbitrary automorphism of $F_{2}=\langle x,y\rangle$.

Proposition 4.3**.**

The group $\operatorname{Aut}(P_{3})$ is generated by the set $\{\theta,\xi,\eta,\phi\}$.

Proof.

It is enough to prove that any automorphism $f$ of $P_{3}$ can be expressed in terms of these four automorphisms. Since $Z(P_{3})=\langle z\rangle$, it follows that $z^{f}=z^{\pm 1}$. Further, $f$ induces an automorphism of $P_{3}/Z(P_{3})\cong F_{2}$. Hence $x^{f}\equiv x^{\psi}\mod Z(P_{3})$ and $y^{f}\equiv y^{\psi}\mod Z(P_{3})$ for some $\psi\in\operatorname{Aut}(F_{2})$. Define $\phi:P_{3}\to P_{3}$ by

[TABLE]

Then the automorphism $f\phi^{-1}$ induce identity on $P_{3}/Z(P_{3})$. Hence

[TABLE]

for some $\alpha,\beta\in\mathbb{Z}$. Hence $f=\theta\xi^{\alpha}\eta^{\beta}\phi$. This proves the lemma. ∎

Let $x\mapsto p(x,y)$ and $y\mapsto q(x,y)$ be an arbitrary automorphism of $F_{2}=\langle x,y\rangle$.

Lemma 4.4**.**

The group $\operatorname{Aut}(P_{3})$ has following defining relations:

1. (1)

$\theta^{2}=1$** 2. (2)

$\xi\eta=\eta\xi$** 3. (3)

$\theta\xi\theta=\xi^{-1}$** 4. (4)

$\theta\eta\theta=\eta^{-1}$** 5. (5)

$\theta\phi=\phi\theta$* for all $\phi\in\operatorname{Aut}(F_{2})$* 6. (6)

$\xi\phi\xi^{-1}=\xi^{1-\log_{x}(p)}\eta^{-\log_{x}(q)}\phi$** 7. (7)

$\eta\phi\eta^{-1}=\xi^{-\log_{y}(p)}\eta^{1-\log_{y}(q)}\phi$.

Proof.

(1)-(5) are evident from the definitions of the automorphisms. For (6), consider

[TABLE]

Similarly, for $y$, we obtain

[TABLE]

The identity (7) follows analogously. ∎

Remark 4.5**.**

Conditions (6) and (7) can be rewritten in the following form:

1. (6)́

$\phi\xi\phi^{-1}=\xi^{\log_{x}(p)}\eta^{\log_{x}(q)}$. 2. (7)́

$\phi\eta\phi^{-1}=\xi^{\log_{y}(p)}\eta^{\log_{y}(q)}$.

The above relations yield the following lemma.

Lemma 4.6**.**

$\langle\xi,\eta\rangle$* is a normal subgroup of $\operatorname{Aut}(P_{3})$ and $\operatorname{Aut}(P_{3})/\langle\xi,\eta\rangle\cong\mathbb{Z}_{2}\times\operatorname{Aut}(F_{2})$.*

Recall that $P_{3}=\mathbb{Z}\times F_{2}$. In this case, $\operatorname{Aut}(P_{3})=\operatorname{Aut}_{c}(P_{3})\times\operatorname{Aut}(F_{2})$, where $\operatorname{Aut}_{c}(P_{3})=\mathbb{Z}^{2}\rtimes\mathbb{Z}_{2}$, generated by $\psi,\phi_{1,3},\phi_{2,3}$ with $\psi^{2}=1$ and $\psi\phi_{i,j}\psi=\phi_{i,j}^{-1}$. The generators are given by

[TABLE]

[TABLE]

and

[TABLE]

Note that $F_{2}=\overline{P}_{3}=P_{3}/Z(P_{3})=P_{3}/\langle A_{1,2}A_{1,3}A_{2,3}\rangle=\langle A_{1,3},A_{2,3}\rangle$. The group $\operatorname{Aut}(F_{2})$ admits the following presentation

[TABLE]

where the automorphisms are given by

[TABLE]

[TABLE]

and

[TABLE]

The lifts of these automorphisms to automorphisms of $P_{3}$ fixing the $A_{1,2}A_{1,3}A_{2,3}$ are given by setting

[TABLE]

Thus, we have proved the following.

Proposition 4.7**.**

The group $\operatorname{Aut}(P_{3})$ is generated by the set $\{\rho,\sigma,\nu,\psi,\phi_{1,3},\phi_{2,3}\}.$

5. Extension and lifting problem for $1\to U_{n}\to P_{n}\to P_{n-1}\to 1$

In this section, we deal the extension and lifting problem for the exact sequence

[TABLE]

The cases $n=1,2$ are vacuous. For $n=3$, we have $P_{3}\cong U_{3}\times P_{2}$, and hence each automorphism of $U_{3}$ and $P_{2}$ can be extended to an automorphism of $P_{3}$.

We deal the case $n=4$ in the rest of this section. Recall that, by definition, $P_{4}=\langle A_{1,2},A_{1,3},A_{2,3},A_{1,4},A_{2,4},A_{3,4}\rangle$ and $P_{3}=\langle A_{1,2},A_{1,3},A_{2,3}\rangle$. Further, $U_{4}=\langle A_{1,4},A_{2,4},A_{3,4}\rangle$ is normal in $P_{4}$ and $P_{4}/U_{4}\cong P_{3}$. The main result of this section is the following theorem.

Theorem 5.1**.**

There exists an automorphism of $P_{3}$ which cannot be lifted to an automorphism of $P_{4}$.

The theorem will be proved via the following sequence of lemmas.

Lemma 5.2**.**

Let $G$ be a group and $\phi\in\operatorname{Aut}(G)$. For $g\in G$, let $\operatorname{C}_{G}(g)=\{x\in G~{}|~{}gx=xg\}$. Then $\operatorname{C}_{G}(g^{\phi})=\operatorname{C}_{G}(g)^{\phi}$.

Proof.

Let $x\in\operatorname{C}_{G}(g)$. Then $xg=gx$ implies that $x^{\phi}g^{\phi}=g^{\phi}x^{\phi}$. This further implies $x^{\phi}\in\operatorname{C}_{G}(g^{\phi})$. Hence $\operatorname{C}_{G}(g)^{\phi}\subseteq\operatorname{C}_{G}(g^{\phi})$. The converse is also obvious. ∎

Lemma 5.3**.**

Let $F_{n}=\langle x_{1},\dots,x_{n}\rangle$, and $\phi\in\operatorname{Aut}(F_{n})$ such that

[TABLE]

Then $Fix(\phi)=F_{n-1}=\langle x_{1},\dots,x_{n-1}\rangle$.

Proof.

Let $g\in Fix(\phi)$. Then we can write $g=w_{1}x_{n}^{\alpha_{1}}\cdots w_{m}x_{n}^{\alpha_{m}}w_{m+1}$ for some $\alpha_{1},\dots,\alpha_{m}\in\mathbb{Z}$ and $w_{1},\dots,w_{m},w_{m+1}\in F_{n-1}$. We can assume that $w_{m+1}=1$ and $\alpha_{m}\neq 0$. Applying $\phi$ gives

[TABLE]

Note that the elements $ww_{i}w^{-1}\neq 1$ and we have equality in the free product $F_{n}=F_{n-1}\star\langle x_{n}\rangle$. This implies that $m=1$ and $\alpha_{m}=0$. Hence $g\in F_{n-1}$. ∎

We also need the well-known conjugation rules in $P_{4}$. Consider the automorphism of $P_{3}$ of the following form

[TABLE]

Lemma 5.4**.**

$Fix(\widehat{A_{1,3}})=\langle A_{1,4}A_{3,4},A_{1,4}A_{2,4}A_{3,4}\rangle$.

Proof.

Consider

[TABLE]

Clearly, $(A_{1,4}A_{3,4})^{A_{1,3}}=A_{1,4}A_{3,4}$. Note that

[TABLE]

In these generators, we have

[TABLE]

From Lemma 5.3, we obtain $Fix(A_{1,3})=\langle A_{1,4}A_{3,4},A_{1,4}A_{2,4}A_{3,4}\rangle$. ∎

Lemma 5.5**.**

$Fix(\widehat{A_{1,3}A_{2,3}})=\langle A_{1,4}A_{2,4}A_{3,4}\rangle$.

Proof.

First, note that $A_{1,4}A_{2,4}A_{3,4}\in Fix(A_{1,3}A_{2,3})$. Now, we determine action of $A_{1,3}A_{2,3}$ on $U_{4}$. We have

[TABLE]

and

[TABLE]

Also, we have $(A_{1,4}A_{2,4}A_{3,4})^{A_{1,3}A_{2,3}}=A_{1,4}A_{2,4}A_{3,4}$. If we set $x=A_{1,4}A_{2,4}A_{3,4}$, $y=A_{3,4}$ and $z=A_{1,4}$, then

[TABLE]

Let $w\in Fix(A_{1,3}A_{2,3})$. Then $w=w_{1}z^{\alpha_{1}}w_{2}z^{\alpha_{2}}\cdots w_{m}z^{\alpha_{m}}w_{m+1}$, where $w_{1},w_{2},\dots,w_{m+1}\in\langle y,x\rangle$. Now

[TABLE]

This implies $(xy)^{-1}w_{m}^{x}=w_{m+1}$, which is a contradiction if $\alpha_{m}\neq 0$. Hence $\alpha_{1}=\cdots=\alpha_{m}=0$ and $w=x^{\alpha_{1}}y^{\beta_{1}}\cdots x^{\alpha_{n}}y^{\beta_{n}}$. Now, by Lemma 5.3, $w=x^{\alpha}$ for some $\alpha$. This completes the proof of the lemma. ∎

**Proof of Theorem 5.1. ** If we lift the automorphism $\phi$ to $P_{4}$, then $Fix(\widehat{A_{1,3}})\cong F_{2}$ and $Fix(\widehat{A_{1,3}A_{2,3}})\cong\mathbb{Z}$, which is a contradiction. Hence $\phi$ cannot be lifted to $P_{4}$. $\Box$

In the reverse direction, we have the following.

Theorem 5.6**.**

There exists an automorphism of $U_{4}$ which cannot be extended to an automorphism of $P_{4}$.

Proof.

Consider the automorphism of $U_{4}$ of the following form

[TABLE]

Suppose that we can extend $\phi$ to an automorphism of $P_{4}$. Then $\phi$ induces an automorphism of $P_{3}$. Recall that $Z(P_{3})=\langle A_{1,2}A_{1,3}A_{2,3}\rangle$. Then $(A_{1,2}A_{1,3}A_{2,3})^{\phi}=(A_{1,2}A_{1,3}A_{2,3})^{\pm 1}$. Also, $A_{1,2}A_{1,3}A_{2,3}$ acts on $U_{4}$ by conjugating $A_{1,4}A_{2,4}A_{3,4}$. Hence $(A_{1,4}A_{2,4}A_{3,4})^{A_{1,2}A_{1,3}A_{2,3}}=A_{1,4}A_{2,4}A_{3,4}$. Applying $\phi$, we get

[TABLE]

This further implies $(A_{1,4}A_{2,4}^{2}A_{3,4})^{(A_{1,4}A_{2,4}A_{3,4})^{\pm 1}}=A_{1,4}A_{2,4}^{2}A_{3,4}$, which is a contradiction. Hence $\phi$ cannot be extended to an automorphism of $P_{4}$. ∎

Question 3**.**

What can we say about the lifting and extension problem for the exact sequence

[TABLE]

for $n\geq 5$.

6. Extension and lifting problem for $1\to P_{n}\to B_{n}\to S_{n}\to 1$

In this section, we investigate the extension and lifting problem for the well-known short exact sequence

[TABLE]

The following is a straightforward observation.

Lemma 6.1**.**

Let $1\to K\to G\to H\to 1$ be a short exact sequence of groups. Then every $\phi\in\operatorname{Inn}(H)$ has a lift in $\operatorname{Aut}(G)$ and every $\psi\in\operatorname{Inn}(K)$ has an extension in $\operatorname{Aut}(G)$.

We now prove the following.

Proposition 6.2**.**

If $n\neq 6$, then any automorphism $\phi\in\operatorname{Aut}(S_{n})$ can be lifted to an automorphism of $B_{n}$. Further, the non-inner automorphism of $S_{6}$ cannot be lifted to an automorphism of $B_{6}$.

Proof.

We know that if $n\neq 2$ or 6, then $\operatorname{Aut}(S_{n})=\operatorname{Inn}(S_{n})$, and hence the result follows from Lemma 6.1. For $n=2$, we have $B_{2}=\langle\sigma_{1}\rangle$, $P_{2}=\langle\sigma_{1}^{2}\rangle$ and $S_{2}=\mathbb{Z}_{2}$. Note that $\operatorname{Aut}(S_{2})=1$, and the identity automorphism is obviously liftable to $B_{2}$. On the other hand, $\operatorname{Aut}(P_{2})=\mathbb{Z}_{2}$, say, generated by $\psi$. Then $\psi(\sigma_{1}^{2})=\sigma_{1}^{-2}$. Define $\psi^{\circ}:B_{2}\to B_{2}$ by $\psi^{\circ}(\sigma_{1})=\sigma_{1}^{-1}$. Then $\psi^{\circ}$ is an extension of $\psi$.

Finally, let $n=6$ and $\phi\in\operatorname{Aut}(S_{6})\setminus\operatorname{Inn}(S_{6})$. Since $\operatorname{Aut}(S_{6})=\operatorname{Inn}(S_{6})\rtimes\mathbb{Z}_{2}$, we have $\phi^{2}=1$. If $\tilde{\phi}$ is a lift of $\phi$, then $\tilde{\phi}=\hat{g}$ for some $g\in B_{6}$ or $\tilde{\phi}=\hat{g}\tau$, where $\tau(\sigma_{i})=\sigma_{i}^{-1}$. Here $\hat{g}$ is the inner automorphism of $B_{6}$ induced by $g$. If $\tilde{\phi}=\hat{g}$, then applying on the generators yield

[TABLE]

for all $1\leq i\leq n-1$. This implies that $\phi=\widehat{g^{\pi}}$, a contradiction.

Now suppose that $\tilde{\phi}=\hat{g}\tau$. Then

[TABLE]

for all $1\leq i\leq n-1$. This implies that $\phi=\widehat{\tau(g)^{\pi}}$, which is again a contradiction. ∎

Recall that $P_{n}\cong Z(P_{n})\times P_{n}/Z(P_{n})$, where $Z(P_{n})=\langle z_{n}\rangle$ is infinite cyclic. Further, the group of central automorphisms of $P_{n}$ is given by

[TABLE]

where $tv^{\circ}(P_{n})=\mathbb{Z}^{N}$ with $N=\binom{n}{2}-1$ and $\mathbb{Z}_{2}=\langle\theta_{0}\rangle$. Here $z_{n}^{\theta_{0}}=z_{n}^{-1}$ and it acts trivially on $P_{n}/Z(P_{n})$.

Proposition 6.3**.**

No non-trivial element of $tv^{\circ}(P_{n})$ can be extended to an automorphism of $B_{n}$.

Proof.

First observe that

[TABLE]

is a finite group. Let $\phi^{\circ}\in tv^{\circ}(P_{n})$ be extendable to an automorphism $\phi$ of $B_{n}$. Then there exists an integer $m$ such that $\phi^{m}\in\operatorname{Inn}(P_{n})$, and hence $\phi^{m}|_{P_{n}}={\phi^{\circ}}^{m}|_{P_{n}}$. This implies $\phi^{m}\in tv^{\circ}(P_{n})\cap\operatorname{Inn}(P_{n})=\{1\}$, and hence ${\phi^{\circ}}^{m}=1$. But the group $tv^{\circ}(P_{n})$ is free abelian, and hence does not have any non-trivial element of finite order. Therefore $\phi^{\circ}=1$. ∎

Lemma 6.4**.**

Let $\tau\in\operatorname{Aut}(B_{n})$ be the automorphism given by $\tau(\sigma_{i})=\sigma_{i}^{-1}$ for $1\leq i\leq n-1$. Then $z_{n}^{\tau}=z_{n}^{-1}$.

Proof.

Recall that $z_{n}=(\sigma_{1}\cdots\sigma_{n-1})^{n}$. Thus $z_{n}^{\tau}=(\sigma_{1}^{-1}\cdots\sigma_{n-1}^{-1})^{n}$. Since $Z(B_{n})=\langle z_{n}\rangle\cong\mathbb{Z}$, we have $z_{n}^{\tau}=z_{n}^{\pm 1}$. Note that $B_{n}/B_{n}^{\prime}$ is infinite cyclic generated by $\sigma_{1}B_{n}^{\prime}$. If $z_{n}^{\tau}=z_{n}$, then reading this equation modulo $B_{n}^{\prime}$ gives $\sigma_{1}^{(n-1)n}\equiv\sigma_{1}^{-(n-1)n}\mod B_{n}^{\prime}$. Equivalently, $\sigma_{1}^{2(n-1)n}\equiv 1\mod B_{n}^{\prime}$, which is a contradiction as $B_{n}/B_{n}^{\prime}$ has no element of finite order. Hence $z_{n}^{\tau}=z_{n}^{-1}$. ∎

Proposition 6.5**.**

$\theta_{0}$* cannot be extended to an automorphism of $B_{n}$.*

Proof.

Suppose that $\theta\in\operatorname{Aut}(B_{n})$ is an extension of $\theta_{0}$. Then $z_{n}^{\theta}=z_{n}^{\theta_{0}}=z_{n}^{-1}$. Recall that $\operatorname{Aut}(B_{n})\cong\operatorname{Inn}(B_{n})\rtimes\langle\tau\rangle$. Since for each $\psi\in\operatorname{Inn}(B_{n})$, we have $z_{n}^{\psi}=z_{n}$, it follows that $\theta=\tau\psi$ for some $\psi\in\operatorname{Inn}(B_{n})$. It follows from the relations in $B_{n}$ that $\psi\in\operatorname{Inn}(B_{n})$ induces a permutation of $P_{n}/P_{n}^{\prime}$. On the other hand, $\tau$ induces inversion in $P_{n}/P_{n}^{\prime}$, more precisely, $A_{i,j}^{\tau}\equiv A_{i,j}^{-1}\mod P_{n}^{\prime}$ for all $i<j$. We can assume that

[TABLE]

Then we have

[TABLE]

for some $1\leq i<j\leq n$, which is a contradiction. This proves the proposition. ∎

Finally, we prove the main result of this section.

Theorem 6.6**.**

No non-trivial element of $\operatorname{Aut}_{c}(P_{n})$ can be extended to an automorphism of $B_{n}$.

Proof.

Let $\phi=\alpha\tau^{\epsilon}\in\operatorname{Aut}_{c}(P_{n})$, where $\alpha\in tv^{\circ}(P_{n})$ and $\epsilon=0,1$. In view of Propositions 6.3 and 6.5, we can assume that $\alpha\neq 1$ and $\epsilon=1$. Note that $\phi^{2}=(\alpha\tau)^{2}=\alpha^{2}\neq 1$ since $tv^{\circ}(P_{n})$ is free abelian. By Proposition 6.3, $\alpha^{2}$ cannot be extended to an automorphism of $B_{n}$. Therefore $\phi^{2}$, and hence $\phi$ cannot be extended to an automorphism of $B_{n}$. ∎

The following question remains.

Question 4**.**

Which non-central automorphisms of $P_{n}$ can be extended to automorphisms of $B_{n}$?

Acknowledgement**.**

The authors gratefully acknowledge support from the Russian Science Foundation Project No.16-41-02006 and the DST-RSF Project INT/RUS/RSF/P-2.