ON THE POLYNOMIAL AUTOMORPHISMS OF A
GROUP
GÉRARD ENDIMIONI
Abstract. Let A(G) denote the automorphism group of a group
G. A polynomial automorphism of G is an automorphism of the
form x 7→ (v−1
ǫ1v1) . . . (v
). We prove that if G is nilpo-
tent (resp. metabelian), then so is the subgroup of A(G) generated
by all polynomial automorphisms.
1. Introduction and main results
Let G be a group. We shall write A(G) for the automorphism group
of G. According to Schweigert [10], we say that an element f ∈ A(G)
is a polynomial automorphism of G if there exist integers ǫ1, . . . , ǫm ∈ Z
and elements u0, . . . , um ∈ G such that
f(x) = u0x
ǫ1u1 . . . um−1x
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(x) = (v−1
xǫ1v1) . . . (v
ǫmvm).
We shall write P0(G) for the set of polynomial automorphisms of G.
Actually, Schweigert defines 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. For instance, in the additive group of rational
numbers, the set of polynomial automorphisms forms a monoid with
respect to the operation of functional composition, which is isomorphic
to the multiplicative monoid Z \ {0}.
1991 Mathematics Subject Classification. 20F28, 20F16, 20F18.
Key words and phrases. polynomial automorphism, metabelian group, nilpotent
group, IA-automorphism.
http://arxiv.org/abs/0704.0500v1
2 GÉRARD ENDIMIONI
In this paper, we shall consider the subgroup P(G) = 〈P0(G)〉 of
A(G), generated by all polynomial automorphisms ofG. Hence P0(G) =
P(G) when G is finite, but for example P(G) is distinct from P0(G)
when G is the additive group of rational numbers (note that P(G) =
A(G) in this last case).
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(G)E P(G)E A(G),
where I(G) is the group of inner automorphisms of G. Also P(G)
contains the group of invertible elements of the monoid P0(G). It is
worth noting that there exist finite groups G such that the quotient
P(G)/I(G) is not soluble [7].
If G is abelian, each polynomial automorphism is of the form x 7→ xǫ,
and so P(G) is abelian. When G is a finite nilpotent group of class
k ≥ 2, it is proved in [4] that P(G) is nilpotent of class k − 1 (see also
[10, Satz 3.5]). We show here that this result remains true when G is
infinite.
Theorem 1.1. Let G be a nilpotent group of class k ≥ 2. Then P(G)
is nilpotent of class k − 1.
Notice that conversely, if P(G) is nilpotent, then so is G since P(G)
contains the group of inner automorphisms.
When G is metabelian, it seems that nothing is known about P(G),
even in the context of finite groups. In this paper, we shall prove the
following.
Theorem 1.2. Let G be a metabelian group. Then the group P(G) is
itself metabelian.
In Section 3, we shall interpret a result of C. K. Gupta as a very
particular case of this theorem (see Corollary 3.1 below).
2. Proofs
As usual, in a group G, the commutator of two elements x, y is de-
fined by [x, y] = x−1y−1xy. Instead of [[x, y], z], we shall write [x, y, z].
We denote by [G,G] the derived subgroup of G.
POLYNOMIAL AUTOMORPHISMS 3
Lemma 2.1. Let f, g be two functions over a group G, respectively
defined by the relations
f(x) = (v−1
xǫ1v1) . . . (v
ǫmvm),
g(x) = (w−1
xη1w1) . . . (w
ηnwn)
(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(g(t)) =
tǫiηj [tǫiηj , vi][t
ǫiηj , wj][t
ǫiηj , wj, vi]
(notice that in this product, the order of the factors is of no conse-
quence).
Proof. Using the fact that any two conjugates of t commute, we can
write
f(g(t)) =
w−1j t
v−1i w
ǫiηjwjvi
tǫiηj [tǫiηj , wjvi].
We conclude thanks to the relation [x, yz] = [x, z][x, y][x, y, z]. �
In a nilpotent group G of class ≤ 2, two conjugates of any element
t ∈ G commute. Therefore, as an immediate consequence of Lemma
2.1, we observe that any two polynomial automorphisms of G commute.
Since these automorphisms generate P(G), we obtain:
Corollary 2.1. If G is a nilpotent group of class ≤ 2, then P(G) is
abelian.
We are now ready to prove our first theorem.
Proof of Theorem 1.1. Since P(G) contains I(G) (which is nilpotent of
class k − 1 exactly), it suffices to show that P(G) is nilpotent of class
at most k − 1. We argue by induction on the nilpotency class k of
4 GÉRARD ENDIMIONI
G. The case k = 2 follows from Corollary 2.1. Now suppose that our
theorem is proved for an integer k ≥ 2 and consider a nilpotent group
G of class k + 1. Denote by ζ(G) the centre of G. One can define a
homomorphism Θ : P(G) → A(G/ζ(G)), where for each f ∈ P(G),
Θ(f) is the automorphism induced by f in G/ζ(G). Clearly, if f is a
polynomial automorphism of G, then Θ(f) is a polynomial automor-
phism of G/ζ(G). Hence Θ(P(G)) is a subgroup of P(G/ζ(G)), and so,
by induction, is nilpotent of class at most k − 1. Since Θ(P(G)) and
P(G)/ kerΘ are isomorphic, it suffices to show that ker Θ is included
in the centre of P(G) and the theorem is proved. For that, consider
an element g ∈ ker Θ and put w(x) = x−1g(x) for any x in G. Thus
g(x) = xw(x) and w(x) belongs to ζ(G) for all x ∈ G. Notice that w
defines a homomorphism of G into ζ(G) since
w(xy) = y−1x−1g(x)g(y) = y−1w(x)g(y) = w(x)w(y).
In order to show that g belongs to the centre of P(G), it suffices to
verify that g commutes with any polynomial automorphism f of G.
Suppose that f is defined by the relation
f(x) = (v−1
xǫ1v1) . . . (v
ǫmvm).
We have easily
f(g(x)) = f(xw(x)) = f(x)f(w(x)) = f(x)w(x)ǫ,
where ǫ = ǫ1 + · · ·+ ǫm. In the same way, by using the fact that w is
a homomorphism, we can write
g(f(x)) = f(x)w(f(x))
= f(x)(w(v1)
−1w(x)ǫ1w(v1)) . . . (w(vm)
−1w(x)ǫmw(vm)),
whence g(f(x)) = f(x)w(x)ǫ. Thus g and f commute, as required, and
the result follows. �
Now we undertake the proof of our second theorem. First we need
the following result, which is well known and easy to prove (see for
example [8, Lemma 34.51] or [9, Part 2, p. 64]).
Lemma 2.2. In a metabelian group G, if t is an element of the derived
subgroup [G,G], we have the relation [t, x, y] = [t, y, x] for all x, y ∈ G.
POLYNOMIAL AUTOMORPHISMS 5
We arrive to the key lemma in the proof of Theorem 1.2. This
lemma shows that when G is metabelian, any element h ∈ [P(G),P(G)]
operates trivially on [G,G] and on G/[G,G].
Lemma 2.3. Let G be a metabelian group. Suppose that h is an ele-
ment of the derived subgroup [P(G),P(G)]. Then
(i) h(t) = t for all t ∈ [G,G];
(ii) x−1h(x) belongs to [G,G] for all x ∈ G.
Proof. (i) Consider the homomorphism Φ : P(G) → A([G,G]) defined
like this: for any f ∈ P(G), Φ(f) is the restriction of f to [G,G]. We
must show that ker Φ contains [P(G),P(G)]. For that, first notice that
any two conjugates of t ∈ [G,G] commute since G is metabelian. Now
we apply Lemma 2.1. If f and g are polynomial automorphisms of G
defined as in this lemma, we obtain the equalities
f(g(t)) =
tǫiηj [tǫiηj , vi][t
ǫiηj , wj][t
ǫiηj , wj, vi],
g(f(t)) =
tǫiηj [tǫiηj , vi][t
ǫiηj , wj][t
ǫiηj , vj, wi],
and so, by Lemma 2.2, f(g(t)) = g(f(t)) for all t ∈ [G,G]. It follows
that [f, g] belongs to ker Φ. In other words, the images of f and g in
P(G)/ ker Φ commute. Since P(G)/ ker Φ is generated by the images
of the polynomial automorphisms, this quotient is abelian. It follows
that ker Φ contains [P(G),P(G)], as desired.
(ii) Here, we consider the homomorphism Ψ : P(G) → A(G/[G,G]),
where for any f ∈ P(G), Ψ(f) is the automorphism induced inG/[G,G]
by f . Since a polynomial automorphism of G induces in G/[G,G]
a polynomial automorphism of G/[G,G], Ψ(P(G)) is a subgroup of
P(G/[G,G]). But P(G/[G,G]) is abelian (see for instance Corollary 2.1
above) and Ψ(P(G)) is isomorphic to P(G)/ kerΨ. Hence P(G)/ ker Ψ
is abelian. Consequently, kerΨ contains [P(G),P(G)] and the result
follows. �
Proof of Theorem 1.2. Let f, g be two elements of [P(G),P(G)]. For
any x ∈ G, put v(x) = x−1f(x) and w(x) = x−1g(x). By Lemma 2.3,
6 GÉRARD ENDIMIONI
v(x) and w(x) belong to [G,G]. Applying again Lemma 2.3, we can
write
f(g(x) = f(xw(x)) = f(x)f(w(x)) = xv(x)w(x).
In the same way, we have g(f(x)) = xw(x)v(x) = xv(x)w(x). It follows
that f and g commute. Thus [P(G),P(G)] is abelian, and so P(G) is
metabelian. �
3. IA-automorphisms of two-generator metabelian groups
By way of illustration, we apply Theorem 1.2 to IA-automorphisms
of a two-generator metabelian group. We recall that an automorphism
of a group G is said to be an IA-automorphism if it induces the iden-
tity automorphism on G/[G,G]. In a free metabelian group of rank
2, each IA-automorphism is inner [1], and so is a polynomial automor-
phism. It turns out that in any two-generator metabelian group, each
IA-automorphism is polynomial. This result is implicit in [2] with a
different terminology. For convenience, we give a proof since this one
is short and elementary.
Proposition 3.1. Each IA-automorphism of a two-generator metabelian
group is polynomial.
To prove this proposition, we shall use the following result.
Lemma 3.1. In a metabelian group G, each function ϕ of the form
x 7→ ϕ(x) = x[x, v1]
η1 . . . [x, vn]
ηn (vi ∈ G, ηi ∈ Z)
is an endomorphism.
Proof. Thanks to the relation [xy, z] = y−1[x, z]y[y, z], we get
ϕ(xy) = xy
y−1[x, vi]y[y, vi]
POLYNOMIAL AUTOMORPHISMS 7
But since the derived subgroup of G is abelian, we can write
ϕ(xy) = xy
y−1[x, vi]
[y, vi]
[x, vi]
[y, vi]
= ϕ(x)ϕ(y),
as required. �
Proof of Proposition 3.1. Suppose thatG is a two-generator metabelian
group generated by a and b. If f is an IA-automorphism of G, we have
f(a) = av and f(b) = bw, where v and w belong to the derived sub-
group [G,G]. Now notice that [G,G] is the normal closure of [a, b].
Therefore, [G,G] is generated by [a, b] and the elements of the form
[a, b, u], with u ∈ G. Hence v and w can be written in the form
v = [a, b]α
[a, b, vi]
w = [a, b]β
[a, b, wi]
where α, β, λ1, . . . , λn, µ1, . . . , µn are integers (possibly equal to 0). By
using the relation [x, y, z] = [x, y]−1[x, z]−1[x, yz], we obtain
v = [a, b]α−λ
[a, vi]
−λi [a, bvi]
w = [a, b]β−µ
[a, wi]
−µi [a, bwi]
where λ = λ1 + · · ·+ λn and µ = µ1 + · · ·+ µn. Now put
ϕ(x) = x[x, b]α−λ[x, a]µ−β
[x, vi]
−λi[x, bvi]
λi[x, wi]
µi [x, awi]
−µi .
By Lemma 3.1, ϕ is an endomorphism of G. Moreover, we have
ϕ(a) = a[a, b]α−λ
[a, vi]
−λi[a, bvi]
λi = av = f(a)
8 GÉRARD ENDIMIONI
since [a, wi] = [a, awi].
In the same way, we get
ϕ(b) = b[a, b]β−µ
[b, wi]
µi [b, awi]
−µi .
By using the identity [a, wi]
−1[a, bwi] = [b, awi]
−1[b, wi] (valid in any
group), we obtain
ϕ(b) = b[a, b]β−µ
[a, wi]
−µi [a, bwi]
µi = bw = f(b).
Thus f = ϕ and the proof is complete. �
We remark that Proposition 3.1 cannot be extended to three-generator
metabelian groups. For example, in the free metabelian group of rank
3 freely generated by a, b, c, consider the IA-automorphism f defined
by f(a) = a, f(b) = b and f(c) = c[a, b]. Suppose that f is polynomial.
Since [a, b] = c−1f(c), the commutator [a, b] would be in the normal
closure of c, hence would be a product of conjugates of c±1. Substi-
tuting 1 for c in this expression gives then [a, b] = 1, a contradiction.
Therefore f is an IA-automorphism which is not polynomial.
As a consequence of Theorem 1.2 and Proposition 3.1, we obtain an
alternative proof of a result due to C. K. Gupta [6] (see also [3]).
Corollary 3.1 ([6]). In a two-generator metabelian group, the group
of IA-automorphisms is metabelian.
Let Md denote the free metabelian group of rank d. By a result of
Bachmuth [1], if d ≥ 3, the group of IA-automorphisms of Md contains
a subgroup which is (absolutely) free of rank d. Thus Corollary 3.1
fails in a d-generator metabelian group when d ≥ 3. Also Bachmuth’s
result shows once again that the group of IA-automorphisms of Md is
not included in P(Md) (if d ≥ 3), since P(Md) is metabelian.
In conclusion we mention that the metabelian groups constitute an
important source of polynomial endomorphisms and automorphisms.
Indeed, by Lemma 3.1, each function of the form
x 7→ x[x, v1]
η1 . . . [x, vn]
ηn (vi ∈ G, ηi ∈ Z)
POLYNOMIAL AUTOMORPHISMS 9
is an endomorphism in a metabelian group G. Besides, when G is
metabelian and nilpotent, such an endomorphism is an automorphism
since in a nilpotent group, every function of the form
x 7→ u0x
ǫ1u1 . . . um−1x
ǫmum (ui ∈ G, ǫi ∈ Z)
is a bijection if ǫ1 + · · ·+ ǫm = ±1 (see [5, Theorem 1]).
References
[1] S. Bachmuth, Automorphisms of free metabelian groups, Trans. Amer. Math.
Soc. 118 (1965), 93–104.
[2] A. Caranti and C. M. Scoppola, Endomorphisms of two-generated metabelian
groups that induce the identity modulo the derived subgroup, Arch. Math. 56
(1991), 218–227.
[3] F. Catino and M. M. Miccoli, A note on IA-endomorphisms of two-generated
metabelian groups, Rend. Sem. Mat. Univ. Padova 96 (1996), 99–104.
[4] G. Corsi Tani and M. F. Rinaldi Bonafede, Polynomial automorphisms in
nilpotent finite groups, Boll. U.M.I. 5 (1986), 285–292.
[5] G. Endimioni, Applications rationnelles d’un groupe nilpotent, C. R. Acad.
Sci. Paris 314 (1992), 431–434.
[6] C. K. Gupta, IA-automorphisms of two-generator metabelian groups, Arch.
Math. 37 (1981), 106–112.
[7] G. Kowol, Polynomautomorphismen von Gruppen, Arch. Math. 57 (1991),
114–121.
[8] H. Neumann, Varieties of Groups, Springer-Verlag, Berlin (1967).
[9] D. J. S. Robinson, Finiteness Conditions and Generalized Soluble Groups,
Springer-Verlag, Berlin (1972).
[10] D. Schweigert, Polynomautomorphismen auf endlichen Gruppen, Arch. Math.
29 (1977), 34–38.
C.M.I-Université de Provence, 39, rue F. Joliot-Curie, F-13453 Mar-
seille Cedex 13
E-mail address : endimion@gyptis.univ-mrs.fr
	1. Introduction and main results
	2. Proofs
	3. IA-automorphisms of two-generator metabelian groups
	References

  We prove that if a group is nilpotent (resp. metabelian), then so is the
subgroup of its automorphism group generated by all polynomial automorphisms.

Introduction and main results
Let G be a group. We shall write A(G) for the automorphism group
of G. According to Schweigert [10], we say that an element f ∈ A(G)
is a polynomial automorphism of G if there exist integers ǫ1, . . . , ǫm ∈ Z
and elements u0, . . . , um ∈ G such that
f(x) = u0x
ǫ1u1 . . . um−1x
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(x) = (v−1
xǫ1v1) . . . (v
ǫmvm).
We shall write P0(G) for the set of polynomial automorphisms of G.
Actually, Schweigert defines 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. For instance, in the additive group of rational
numbers, the set of polynomial automorphisms forms a monoid with
respect to the operation of functional composition, which is isomorphic
to the multiplicative monoid Z \ {0}.
1991 Mathematics Subject Classification. 20F28, 20F16, 20F18.
Key words and phrases. polynomial automorphism, metabelian group, nilpotent
group, IA-automorphism.
http://arxiv.org/abs/0704.0500v1
2 GÉRARD ENDIMIONI
In this paper, we shall consider the subgroup P(G) = 〈P0(G)〉 of
A(G), generated by all polynomial automorphisms ofG. Hence P0(G) =
P(G) when G is finite, but for example P(G) is distinct from P0(G)
when G is the additive group of rational numbers (note that P(G) =
A(G) in this last case).
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(G)E P(G)E A(G),
where I(G) is the group of inner automorphisms of G. Also P(G)
contains the group of invertible elements of the monoid P0(G). It is
worth noting that there exist finite groups G such that the quotient
P(G)/I(G) is not soluble [7].
If G is abelian, each polynomial automorphism is of the form x 7→ xǫ,
and so P(G) is abelian. When G is a finite nilpotent group of class
k ≥ 2, it is proved in [4] that P(G) is nilpotent of class k − 1 (see also
[10, Satz 3.5]). We show here that this result remains true when G is
infinite.
Theorem 1.1. Let G be a nilpotent group of class k ≥ 2. Then P(G)
is nilpotent of class k − 1.
Notice that conversely, if P(G) is nilpotent, then so is G since P(G)
contains the group of inner automorphisms.
When G is metabelian, it seems that nothing is known about P(G),
even in the context of finite groups. In this paper, we shall prove the
following.
Theorem 1.2. Let G be a metabelian group. Then the group P(G) is
itself metabelian.
In Section 3, we shall interpret a result of C. K. Gupta as a very
particular case of this theorem (see Corollary 3.1 below).
2. Proofs
As usual, in a group G, the commutator of two elements x, y is de-
fined by [x, y] = x−1y−1xy. Instead of [[x, y], z], we shall write [x, y, z].
We denote by [G,G] the derived subgroup of G.
POLYNOMIAL AUTOMORPHISMS 3
Lemma 2.1. Let f, g be two functions over a group G, respectively
defined by the relations
f(x) = (v−1
xǫ1v1) . . . (v
ǫmvm),
g(x) = (w−1
xη1w1) . . . (w
ηnwn)
(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(g(t)) =
tǫiηj [tǫiηj , vi][t
ǫiηj , wj][t
ǫiηj , wj, vi]
(notice that in this product, the order of the factors is of no conse-
quence).
Proof. Using the fact that any two conjugates of t commute, we can
write
f(g(t)) =
w−1j t
v−1i w
ǫiηjwjvi
tǫiηj [tǫiηj , wjvi].
We conclude thanks to the relation [x, yz] = [x, z][x, y][x, y, z]. �
In a nilpotent group G of class ≤ 2, two conjugates of any element
t ∈ G commute. Therefore, as an immediate consequence of Lemma
2.1, we observe that any two polynomial automorphisms of G commute.
Since these automorphisms generate P(G), we obtain:
Corollary 2.1. If G is a nilpotent group of class ≤ 2, then P(G) is
abelian.
We are now ready to prove our first theorem.
Proof of Theorem 1.1. Since P(G) contains I(G) (which is nilpotent of
class k − 1 exactly), it suffices to show that P(G) is nilpotent of class
at most k − 1. We argue by induction on the nilpotency class k of
4 GÉRARD ENDIMIONI
G. The case k = 2 follows from Corollary 2.1. Now suppose that our
theorem is proved for an integer k ≥ 2 and consider a nilpotent group
G of class k + 1. Denote by ζ(G) the centre of G. One can define a
homomorphism Θ : P(G) → A(G/ζ(G)), where for each f ∈ P(G),
Θ(f) is the automorphism induced by f in G/ζ(G). Clearly, if f is a
polynomial automorphism of G, then Θ(f) is a polynomial automor-
phism of G/ζ(G). Hence Θ(P(G)) is a subgroup of P(G/ζ(G)), and so,
by induction, is nilpotent of class at most k − 1. Since Θ(P(G)) and
P(G)/ kerΘ are isomorphic, it suffices to show that ker Θ is included
in the centre of P(G) and the theorem is proved. For that, consider
an element g ∈ ker Θ and put w(x) = x−1g(x) for any x in G. Thus
g(x) = xw(x) and w(x) belongs to ζ(G) for all x ∈ G. Notice that w
defines a homomorphism of G into ζ(G) since
w(xy) = y−1x−1g(x)g(y) = y−1w(x)g(y) = w(x)w(y).
In order to show that g belongs to the centre of P(G), it suffices to
verify that g commutes with any polynomial automorphism f of G.
Suppose that f is defined by the relation
f(x) = (v−1
xǫ1v1) . . . (v
ǫmvm).
We have easily
f(g(x)) = f(xw(x)) = f(x)f(w(x)) = f(x)w(x)ǫ,
where ǫ = ǫ1 + · · ·+ ǫm. In the same way, by using the fact that w is
a homomorphism, we can write
g(f(x)) = f(x)w(f(x))
= f(x)(w(v1)
−1w(x)ǫ1w(v1)) . . . (w(vm)
−1w(x)ǫmw(vm)),
whence g(f(x)) = f(x)w(x)ǫ. Thus g and f commute, as required, and
the result follows. �
Now we undertake the proof of our second theorem. First we need
the following result, which is well known and easy to prove (see for
example [8, Lemma 34.51] or [9, Part 2, p. 64]).
Lemma 2.2. In a metabelian group G, if t is an element of the derived
subgroup [G,G], we have the relation [t, x, y] = [t, y, x] for all x, y ∈ G.
POLYNOMIAL AUTOMORPHISMS 5
We arrive to the key lemma in the proof of Theorem 1.2. This
lemma shows that when G is metabelian, any element h ∈ [P(G),P(G)]
operates trivially on [G,G] and on G/[G,G].
Lemma 2.3. Let G be a metabelian group. Suppose that h is an ele-
ment of the derived subgroup [P(G),P(G)]. Then
(i) h(t) = t for all t ∈ [G,G];
(ii) x−1h(x) belongs to [G,G] for all x ∈ G.
Proof. (i) Consider the homomorphism Φ : P(G) → A([G,G]) defined
like this: for any f ∈ P(G), Φ(f) is the restriction of f to [G,G]. We
must show that ker Φ contains [P(G),P(G)]. For that, first notice that
any two conjugates of t ∈ [G,G] commute since G is metabelian. Now
we apply Lemma 2.1. If f and g are polynomial automorphisms of G
defined as in this lemma, we obtain the equalities
f(g(t)) =
tǫiηj [tǫiηj , vi][t
ǫiηj , wj][t
ǫiηj , wj, vi],
g(f(t)) =
tǫiηj [tǫiηj , vi][t
ǫiηj , wj][t
ǫiηj , vj, wi],
and so, by Lemma 2.2, f(g(t)) = g(f(t)) for all t ∈ [G,G]. It follows
that [f, g] belongs to ker Φ. In other words, the images of f and g in
P(G)/ ker Φ commute. Since P(G)/ ker Φ is generated by the images
of the polynomial automorphisms, this quotient is abelian. It follows
that ker Φ contains [P(G),P(G)], as desired.
(ii) Here, we consider the homomorphism Ψ : P(G) → A(G/[G,G]),
where for any f ∈ P(G), Ψ(f) is the automorphism induced inG/[G,G]
by f . Since a polynomial automorphism of G induces in G/[G,G]
a polynomial automorphism of G/[G,G], Ψ(P(G)) is a subgroup of
P(G/[G,G]). But P(G/[G,G]) is abelian (see for instance Corollary 2.1
above) and Ψ(P(G)) is isomorphic to P(G)/ kerΨ. Hence P(G)/ ker Ψ
is abelian. Consequently, kerΨ contains [P(G),P(G)] and the result
follows. �
Proof of Theorem 1.2. Let f, g be two elements of [P(G),P(G)]. For
any x ∈ G, put v(x) = x−1f(x) and w(x) = x−1g(x). By Lemma 2.3,
6 GÉRARD ENDIMIONI
v(x) and w(x) belong to [G,G]. Applying again Lemma 2.3, we can
write
f(g(x) = f(xw(x)) = f(x)f(w(x)) = xv(x)w(x).
In the same way, we have g(f(x)) = xw(x)v(x) = xv(x)w(x). It follows
that f and g commute. Thus [P(G),P(G)] is abelian, and so P(G) is
metabelian. �
3. IA-automorphisms of two-generator metabelian groups
By way of illustration, we apply Theorem 1.2 to IA-automorphisms
of a two-generator metabelian group. We recall that an automorphism
of a group G is said to be an IA-automorphism if it induces the iden-
tity automorphism on G/[G,G]. In a free metabelian group of rank
2, each IA-automorphism is inner [1], and so is a polynomial automor-
phism. It turns out that in any two-generator metabelian group, each
IA-automorphism is polynomial. This result is implicit in [2] with a
different terminology. For convenience, we give a proof since this one
is short and elementary.
Proposition 3.1. Each IA-automorphism of a two-generator metabelian
group is polynomial.
To prove this proposition, we shall use the following result.
Lemma 3.1. In a metabelian group G, each function ϕ of the form
x 7→ ϕ(x) = x[x, v1]
η1 . . . [x, vn]
ηn (vi ∈ G, ηi ∈ Z)
is an endomorphism.
Proof. Thanks to the relation [xy, z] = y−1[x, z]y[y, z], we get
ϕ(xy) = xy
y−1[x, vi]y[y, vi]
POLYNOMIAL AUTOMORPHISMS 7
But since the derived subgroup of G is abelian, we can write
ϕ(xy) = xy
y−1[x, vi]
[y, vi]
[x, vi]
[y, vi]
= ϕ(x)ϕ(y),
as required. �
Proof of Proposition 3.1. Suppose thatG is a two-generator metabelian
group generated by a and b. If f is an IA-automorphism of G, we have
f(a) = av and f(b) = bw, where v and w belong to the derived sub-
group [G,G]. Now notice that [G,G] is the normal closure of [a, b].
Therefore, [G,G] is generated by [a, b] and the elements of the form
[a, b, u], with u ∈ G. Hence v and w can be written in the form
v = [a, b]α
[a, b, vi]
w = [a, b]β
[a, b, wi]
where α, β, λ1, . . . , λn, µ1, . . . , µn are integers (possibly equal to 0). By
using the relation [x, y, z] = [x, y]−1[x, z]−1[x, yz], we obtain
v = [a, b]α−λ
[a, vi]
−λi [a, bvi]
w = [a, b]β−µ
[a, wi]
−µi [a, bwi]
where λ = λ1 + · · ·+ λn and µ = µ1 + · · ·+ µn. Now put
ϕ(x) = x[x, b]α−λ[x, a]µ−β
[x, vi]
−λi[x, bvi]
λi[x, wi]
µi [x, awi]
−µi .
By Lemma 3.1, ϕ is an endomorphism of G. Moreover, we have
ϕ(a) = a[a, b]α−λ
[a, vi]
−λi[a, bvi]
λi = av = f(a)
8 GÉRARD ENDIMIONI
since [a, wi] = [a, awi].
In the same way, we get
ϕ(b) = b[a, b]β−µ
[b, wi]
µi [b, awi]
−µi .
By using the identity [a, wi]
−1[a, bwi] = [b, awi]
−1[b, wi] (valid in any
group), we obtain
ϕ(b) = b[a, b]β−µ
[a, wi]
−µi [a, bwi]
µi = bw = f(b).
Thus f = ϕ and the proof is complete. �
We remark that Proposition 3.1 cannot be extended to three-generator
metabelian groups. For example, in the free metabelian group of rank
3 freely generated by a, b, c, consider the IA-automorphism f defined
by f(a) = a, f(b) = b and f(c) = c[a, b]. Suppose that f is polynomial.
Since [a, b] = c−1f(c), the commutator [a, b] would be in the normal
closure of c, hence would be a product of conjugates of c±1. Substi-
tuting 1 for c in this expression gives then [a, b] = 1, a contradiction.
Therefore f is an IA-automorphism which is not polynomial.
As a consequence of Theorem 1.2 and Proposition 3.1, we obtain an
alternative proof of a result due to C. K. Gupta [6] (see also [3]).
Corollary 3.1 ([6]). In a two-generator metabelian group, the group
of IA-automorphisms is metabelian.
Let Md denote the free metabelian group of rank d. By a result of
Bachmuth [1], if d ≥ 3, the group of IA-automorphisms of Md contains
a subgroup which is (absolutely) free of rank d. Thus Corollary 3.1
fails in a d-generator metabelian group when d ≥ 3. Also Bachmuth’s
result shows once again that the group of IA-automorphisms of Md is
not included in P(Md) (if d ≥ 3), since P(Md) is metabelian.
In conclusion we mention that the metabelian groups constitute an
important source of polynomial endomorphisms and automorphisms.
Indeed, by Lemma 3.1, each function of the form
x 7→ x[x, v1]
η1 . . . [x, vn]
ηn (vi ∈ G, ηi ∈ Z)
POLYNOMIAL AUTOMORPHISMS 9
is an endomorphism in a metabelian group G. Besides, when G is
metabelian and nilpotent, such an endomorphism is an automorphism
since in a nilpotent group, every function of the form
x 7→ u0x
ǫ1u1 . . . um−1x
ǫmum (ui ∈ G, ǫi ∈ Z)
is a bijection if ǫ1 + · · ·+ ǫm = ±1 (see [5, Theorem 1]).
References
[1] S. Bachmuth, Automorphisms of free metabelian groups, Trans. Amer. Math.
Soc. 118 (1965), 93–104.
[2] A. Caranti and C. M. Scoppola, Endomorphisms of two-generated metabelian
groups that induce the identity modulo the derived subgroup, Arch. Math. 56
(1991), 218–227.
[3] F. Catino and M. M. Miccoli, A note on IA-endomorphisms of two-generated
metabelian groups, Rend. Sem. Mat. Univ. Padova 96 (1996), 99–104.
[4] G. Corsi Tani and M. F. Rinaldi Bonafede, Polynomial automorphisms in
nilpotent finite groups, Boll. U.M.I. 5 (1986), 285–292.
[5] G. Endimioni, Applications rationnelles d’un groupe nilpotent, C. R. Acad.
Sci. Paris 314 (1992), 431–434.
[6] C. K. Gupta, IA-automorphisms of two-generator metabelian groups, Arch.
Math. 37 (1981), 106–112.
[7] G. Kowol, Polynomautomorphismen von Gruppen, Arch. Math. 57 (1991),
114–121.
[8] H. Neumann, Varieties of Groups, Springer-Verlag, Berlin (1967).
[9] D. J. S. Robinson, Finiteness Conditions and Generalized Soluble Groups,
Springer-Verlag, Berlin (1972).
[10] D. Schweigert, Polynomautomorphismen auf endlichen Gruppen, Arch. Math.
29 (1977), 34–38.
C.M.I-Université de Provence, 39, rue F. Joliot-Curie, F-13453 Mar-
seille Cedex 13
E-mail address : endimion@gyptis.univ-mrs.fr
	1. Introduction and main results
	2. Proofs
	3. IA-automorphisms of two-generator metabelian groups
	References
