The polynomial automorphisms of some certain groups

Abstract

Abstract

Let A(G) denote the automorphism group of a group G. A polynomial automorphism of G is an automorphism of the form $x\mapsto \left(v_1^{-1}{x}^{{\epsilon }_1}{v}_1\right)\dots \left(v_m^{-1}{x}^{{\epsilon }_m}{v}_m\right).$ We shall write P(G)=〈P₀(G)〉 such that P₀(G) is the set of polynomial automorphisms of G. In this paper, we will prove that P₀(D₈)≅V₄ and $P\left(\mathbb{Q}\right)=A\left(\mathbb{Q}\right),$ where $\mathbb{Q}$ is the additive group.

AMS 2010 subject classifications

20D45, 11C08

Introduction

Let G be a group. We shall write A(G) for the automorphism group of G. According to Schweigert [1], we say that an element f∈A(G) is a polynomial automorphism of G if there exist integers ${\epsilon }_1,\dots ,{\epsilon }_m\in \mathbb{Z}$ and elements u₀,...,u _m ∈G such that

$$f\left(x\right)=u_0{x}^{{\epsilon }_1}{u}_1\mathrm{...}{u}_{m-1}{x}^{\mathrm{\epsilon m}}{u}_m,$$

for all x∈G. Since f(1)=1, it is easy to see that f(x) can be expressed as a product of inner automorphisms, that is,

$$f\left(x\right)=\left(v_1^{-1}{x}^{{\epsilon }_1}{v}_1\right)\dots \left(v_m^{-1}{x}^{{\epsilon }_m}{v}_m\right).$$

We shall write P₀(G) for the set of polynomial automorphisms of G. Actually, Schweigert defined a polynomial automorphism in the context of finite groups. In particular, in this context, the set P₀(G) is clearly a subgroup of A(G). On the other hand, this is not necessarily the case when G is infinite.

In this paper, we shall consider the subgroup P(G)=〈P₀(G)〉 of A(G), generated by all polynomial automorphisms of G. Hence, P₀(G)=P(G), when G is finite. For instance, we will prove that P₀(D₈)≅V₄. Instead, P(G) is distinct from P₀(G) when G is the additive group of a rational number; in this case, we will prove that $P\left(\mathbb{Q}\right)=A\left(\mathbb{Q}\right)$, and the set of polynomial automorphisms forms a monoid with respect to the operation of functional composition, which is isomorphic to the multiplicative monoid $\mathbb{Z}\setminus \left\{0\right\}$.

It is easy to verify that P₀(G) is a normal subset of A(G). Thus, P(G) is a normal subgroup of A(G). In addition, we have

$$I\left(G\right)⊴P\left(G\right)⊴A\left(G\right),$$

where I(G) is the group of inner automorphisms of G.

Preliminaries

If G is abelian, each polynomial automorphism is of the form x↦x^ε, and so P(G) is abelian. We show here that if G is a nilpotent group of class k=2, then P(G) is abelian.

Lemma 1. Let f,g be two functions over a group G, respectively defined by the relations

$$f\left(x\right)=\left(v_1^{-1}{x}^{{\epsilon }_1}{v}_1\right)\dots \left(v_m^{-1}{x}^{{\epsilon }_m}{v}_m\right),$$

$$g\left(x\right)=\left(w_1^{-1}{x}^{{\eta }_1}{w}_1\right)\dots \left(w_n^{-1}{x}^{{\eta }_n}{w}_n\right)$$

(we do not suppose that f and g are automorphisms). Let t be an element of G such that any two conjugates of t commute. Then, we have the relation

$$f\left(g\right(t\left)\right)=\prod _{i=1}^m\prod _{j=1}^n{t}^{{\epsilon }_i{\eta }_j}[t^{{\epsilon }_i{\eta }_j},v_i][t^{{\epsilon }_i{\eta }_j},w_j][t^{{\epsilon }_i{\eta }_j},w_j,v_i]$$

(notice that in this product, the order of the factors is of no consequence).

Proof. Using the fact that any two conjugates of t commute, we can write

$$\begin{array}{rrl}f\left(g\right(t\left)\right)& =& {\prod }_{i=1}^m{v}_i^{-1}{\left({\prod }_{j=1}^m{w}_j^{-1}{t}^{{\eta }_j}{w}_j\right)}^{{\epsilon }_i}{v}_i\\ =& {\prod }_{i=1}^m{\prod }_{j=1}^n{v}_i^{-1}{w}_j^{-1}{t}^{{\epsilon }_i{\eta }_j}{w}_j{v}_i\\ =& {\prod }_{i=1}^m{\prod }_{j=1}^n{t}^{{\epsilon }_i{\eta }_j}[t^{{\epsilon }_i{\eta }_j},w_j,v_i].\end{array}$$

We conclude thanks to the relation [x,y z]=[x,z] [x,y][x,y,z]. □

When G is a finite nilpotent group of class ≤2, it is proved in [2], that P(G) is abelian. In a nilpotent group G of class ≤2, two conjugates of any element t∈G commute. Therefore, as an immediate consequence of Lemma 1, we observe that any two polynomial automorphisms of G commute. Since these automorphisms generate P(G), we obtain:

Proposition 1. If G is a nilpotent group of class ≤2, then P(G) is abelian.

Proof. It is enough to show that all generators of P(G) commute. Let G be a nilpotent group of class 1. Then, by [3], we have

$$P_0\left(G\right)=\{f\mid f(x)=x^{\epsilon },\epsilon \in \mathbb{Z}\setminus \{0\left\}\right\}.$$

Now, we consider f(x)=x^ε , g(x)=x^δ. We have

$$\begin{array}{c}f\left(g\right(x\left)\right)=f\left(x^{\delta }\right)=x^{\mathrm{\delta \epsilon }},\\ g\left(f\right(x\left)\right)=g\left(x^{\epsilon }\right)=x^{\mathrm{\epsilon \delta }},\end{array}$$

where $\delta ,\epsilon \in \mathbb{Z}\setminus \left\{0\right\}$. Hence, f(g(x))=g(f(x)).

Let G be a nilpotent group of class 2 and let f,g be two elements of P₀(G) such that

$$\begin{array}{c}f\left(x\right)=\left(v_1^{-1}{x}^{{\epsilon }_1}{v}_1\right)\dots \left(v_m^{-1}{x}^{{\epsilon }_m}{v}_m\right),\\ g\left(x\right)=\left(w_1^{-1}{x}^{{\eta }_1}{w}_1\right)\dots \left(w_n^{-1}{x}^{{\eta }_n}{w}_n\right).\end{array}$$

Then by Lemma 1, for all t ∈ G, we have

$$\begin{array}{c}f\left(g\right(t\left)\right)={\prod }_{i=1}^m{\prod }_{j=1}^n{t}^{{\epsilon }_i{\eta }_j}[t^{{\epsilon }_i{\eta }_j},v_i][t^{{\epsilon }_i{\eta }_j},w_j][t^{{\epsilon }_i{\eta }_j},w_j,v_i],\\ g\left(f\right(t\left)\right)={\prod }_{i=1}^m{\prod }_{j=1}^n{t}^{{\epsilon }_i{\eta }_j}[t^{{\epsilon }_i{\eta }_j},w_j][t^{{\epsilon }_i{\eta }_j},v_i][t^{{\epsilon }_i{\eta }_j},v_i,w_j].\end{array}$$

Since G is a nilpotent group of class 2, γ₃(G)=1. So,

$$[t^{{\epsilon }_i{\eta }_j},w_j,v_i]=[t^{{\epsilon }_i{\eta }_j},v_i,w_j]=1.$$

Therefore, f(g(t))=g(f(t)) and the proof is complete. □

Main results

In this section, we suppose that D₈ is the dihedral group of order 8, V₄ is the Klein 4−group, and $\mathbb{Q}$ is the additive group of a rational number [4, 5]. First, in Theorem 1, we will show that P₀(D₈)≅V₄, and then in Theorem 2, we will prove that$P\left(\mathbb{Q}\right)=A\left(\mathbb{Q}\right).$

Theorem 1. Let D ₈ be the dihedral group of order 8, and let V ₄ be the Klein 4−group. Then,P ₀ (D ₈ )≅V ₄ .

Proof. Since D₈=〈t,s|t²=s⁴=1,(t s)²=1〉, so the eight elements of D₈ are

$$D_8=\{1,s,s^2,s^3,t,\mathrm{ts},t{s}^2,t{s}^3\}.$$

It is straightforward to verify that A u t(D₈)≅D₈. On the other hand, we have D₈/Z(D₈)≅I n n(D₈). Since Z(D₈)=〈s²〉, we have |I n n(D₈)|=4. The order of each non-trivial element of D₈/Z(D₈) is 2, so D₈/Z(D₈)≅V₄; hence, I n n(D₈)≅V₄. Therefore,

$$V_4⊴P_0\left(D_8\right)⊴D_8.$$

Since |D₈:V₄|=2, so P₀(D₈)≅D₈ or V₄. However, D₈ is the nilpotent group of order 2; according to the Proposition 1, P₀(D₈) is abelian group. The result now follows. □

Theorem 2. Let$\mathbb{Q}$be the additive group of rational numbers. Then, the set of polynomial automorphisms forms a monoid with respect to the operation of functional composition, which is isomorphic to the multiplicative monoid$\mathbb{Z}\setminus \left\{0\right\}$. Further, we have$P\left(\mathbb{Q}\right)=A\left(\mathbb{Q}\right)\cong (\mathbb{Q}\setminus \{0\},·).$

Proof. Since $\mathbb{Q}$ is the additive group, so each element of $P_0\left(\mathbb{Q}\right)$ is of the form f(x)=k x for every $x\in \mathbb{Q},$ where $k\in \mathbb{Z}\setminus \left\{0\right\}.$ Now, since $\mathbb{Q}$ is a torsion-free group, so f(x)=k x is the element of $P_0\left(\mathbb{Q}\right)$ for every $k\in \mathbb{Z}\setminus \left\{0\right\}.$ Hence,

$$P_0\left(\mathbb{Q}\right)=\{f:\mathbb{Q}\to \mathbb{Q}|f\left(x\right)=\mathrm{kx},k\in \mathbb{Z}\setminus \left\{0\right\}\}.$$

It can be easily verified that $P_0\left(\mathbb{Q}\right)$ has only two commutative elements. We consider the mapping $\phi :P_0\left(\mathbb{Q}\right)\to \mathbb{Z}\setminus \left\{0\right\}$ defined like this: for any $f\in P_0\left(\mathbb{Q}\right)$, φ(f(x))=k, where $k\in \mathbb{Z}\setminus \left\{0\right\}$ and $x\in \mathrm{\mathbb{Q}.}$ It is easy to see that $P_0\left(\mathbb{Q}\right)\cong (\mathbb{Z}\setminus \{0\},·)$.It is straightforward to verify that every element of $\mathrm{End}\left(\mathbb{Q}\right)$, for every $x\in \mathbb{Q},$ is the form f _t (x)=t x, where t=f _t (1) is the arbitrary element of $\mathbb{Q}$. The mapping $\psi :\mathrm{End}\left(\mathbb{Q}\right)\to \mathbb{Q}$ of the form

$$g_t\mapsto g_t\left(1\right)(g_t\in \mathrm{End}(\mathbb{Q}\left)\right)$$

is an isomorphism.

Since $A\left(\mathbb{Q}\right)\le \mathrm{End}\left(\mathbb{Q}\right)$ and $A\left(\mathbb{Q}\right)$ is the group of invertible elements of $\mathrm{End}\left(\mathbb{Q}\right),$ so we have $A\left(\mathbb{Q}\right)\cong (\mathbb{Q}\setminus \{0\},·).$ Further, we have

$$A\left(\mathbb{Q}\right)=\{f:\mathbb{Q}\to \mathbb{Q}|f\left(x\right)=\mathrm{tx},t\in \mathbb{Q}\setminus \left\{0\right\}\}.$$

It is clear that $P\left(\mathbb{Q}\right)\le A\left(\mathbb{Q}\right).$ Let $f\in A\left(\mathbb{Q}\right)$. Then, there exist $m,n\in \mathbb{Z}\setminus \left\{0\right\}$ such that

$$f\left(x\right)=\frac{m}{n}x$$

for every $x\in \mathbb{Q}$.

Let f₁,f₂ be two elements of $P_0\left(\mathbb{Q}\right),$ respectively defined by the relations

$$f_1\left(x\right)=\mathrm{mx},f_2\left(x\right)=\mathrm{nx.}$$

Then, $f_{1}o{f}_2^{-1}\left(x\right)=\frac{m}{n}x=f\left(x\right).$ Now, since $f_{1}o{f}_2^{-1}$ is the element of $P\left(\mathbb{Q}\right),$ so f(x) is the element of $P\left(\mathbb{Q}\right).$ This complete the proof of Theorem 2. □