Automorphisms fixing every Normal Subgroup of a Nilpotent-by-abelian Group

Automorphisms Fixing Every Normal Subgroup of a Nilpotent-By-Abelian Group.

GEÂRARDENDIMIONI(*)

ABSTRACT- Among other things, we prove that the group of automorphisms fixing every normal subgroup of a (nilpotent of classc)-by-abelian group is (nilpotent of classc)-by-metabelian. In particular, the group of automorphisms fixing every normal subgroup of a metabelian group is soluble of derived length at most 3. An example shows that this bound cannot be improved.

1. Introduction and results.

IfGis a group, we write Aut(G) for the group of all automorphisms ofG.

An automorphism W2Aut(G) is said to be normal if W(H)ˆHfor each normal subgroupHofG. These automorphisms form a normal subgroup of Aut(G), denoted by Autn(G). Obviously, Autn(G) contains the subgroup Inn(G) of all inner automorphisms ofG. These subgroups can coincide, for instance when Gis a nonabelian free group [5]. We have a similar result whenGis a nonabelian free nilpotent group of classcˆ2, but ifc3, then the quotient Autn(G)=Inn(G) is infinite [2]. Also one can find in [3] a de- scription of the normal automorphisms of a free metabelian nilpotent group. In general, the structure of Aut_n(G) can be quite diverse since for an arbitrary finite group F, there is a finite semisimple group Gsuch that Aut_n(G)=Inn(G) has a subgroup isomorphic withF [7]. Nevertheless, as one can expect, it is possible to obtain some information about Autn(G) whenGbelongs to certain classes of groups. For instance, whenGis nil- potent, the group Aut_n(G) is nilpotent-by-abelian [4]. In the same paper, the authors show that Aut_n(G) is polycyclic whenGis. In particular, ifGis a finite soluble group, then so is Autn(G). We do not know if this assertion

(*) Indirizzo dell'A.: C.M.I-Universite de Provence, 39, rue F. Joliot-Curie, F-13453 Marseille Cedex 13, France.

E-mail: endimion@gyptis.univ-mrs.fr

remains true without the term ``finite''. In other words, is Autn(G) soluble whenGis soluble ? Here we respond to this question in the positive whenG is soluble of derived length2, and more generally whenGis nilpotent- by-abelian.

THEOREM1. The group of all normal automorphisms of a (nilpotent of class c)-by-abelian group is (nilpotent of classc)-by-metabelian.

Whencˆ1, we obtain:

COROLLARY1. The group of all normal automorphisms of a metabe- lian group is soluble of derived length at most 3.

As usual, denote byS4the symmetric group of degree 4 and byA4 its alternatingsubgroup. Then Autn(A4)ˆAut(A4) is isomorphic toS4. Since A₄ is metabelian and S₄ is soluble of derived length 3, the bound of the derived length given in the corollary above cannot be improved.

We shall see that the proof of Theorem 1 also leads to the following result.

THEOREM2. Let G be a (nilpotent of class c)-by-abelian group. Sup- pose that its abelianization G=G⁰ is either finite or infinite non-periodic (that is the case for example when G is finitely generated). Then the group of all normal automorphisms of G is virtually (nilpotent of classc)-by- abelian.

In particular, the group of normal automorphisms of a finitely gener- ated metabelian group is virtually metabelian.

Since a supersoluble group is nilpotent-by-abelian and finitely gener- ated, it follows from Theorem 2 that its group of all normal automorphisms is virtually nilpotent-by-abelian. In fact, we have a stronger result.

THEOREM 3. The group of all normal automorphisms of a super- soluble group is finitely generated and nilpotent-by-(finite and super- soluble).

2. Proofs.

LetGbe a group and letG⁰denote its derived subgroup. Clearly, each normal automorphismf 2Autn(G) induces inG=G⁰a normal automorphism

f⁰2Autn(G=G⁰). Consider the homomorphism F:Autn(G)!Autn(G=G⁰) defined byF(f)ˆf⁰ and putKˆkerF. In other words, K is the set of normal IA-automorphisms ofG(recall that an automorphism ofGis said to be anIA-automorphismif it induces the identity automorphism inG=G⁰).

Before to prove Theorem 1, we establish a preliminary result regarding the elements ofK.

LEMMA1. Let k be a positive integer. In a group G, consider an element a2g_k(G⁰), whereg_k(G⁰)denotes the kth term of the lower central series of G⁰. If f and g are normal IA-automorphisms of G (i.e. f;g2K), we have:

(i) for all u2G, f(u ¹au)u ¹f(a)u modg_k‡1(G⁰);

(ii) g ¹f ¹gf(a)a modg_k‡1(G⁰).

PROOF. (i) We havef(u)ˆuu⁰for someu⁰2G⁰, whence f(u ¹au)ˆu^{0 1}u ¹f(a)uu⁰ˆu ¹f(a)u[u ¹f(a)u;u⁰];

where the commutator [x;y] is defined by [x;y]ˆx ¹y ¹xy. Since [u ¹f(a)u;u⁰] belongs tog_k‡1(G⁰), the result follows.

(ii) Since f and g preserve the normal closure of a in G, there exist elements u1;. . .;ur;v1;. . .;vs 2G and integers l1;. . .;lr;m₁;. . .;m_s such that

f(a)ˆY^r

iˆ1

u_i¹a^liuiandg(a)ˆY^s

iˆ1

v_i¹a^mivi:

Notice that in these products, the order of the factors is of no consequence modulog_k‡1(G⁰). Using(i), one can then write

gf(a)Y^r

iˆ1

g(u_i¹a^liu_i) Y^r

iˆ1

u_i¹g(a)^liu_i

Y^r

iˆ1

Y^s

jˆ1

u_i¹v_j¹a^limjv_ju_i

Y^r

iˆ1

Y^s

jˆ1

[vj;ui] ¹v_j¹u_i¹a^limjuivj[vj;ui] modg_k‡1(G⁰):

Since

[v_j;u_i] ¹v_j¹u_i¹a^limju_iv_j[v_j;u_i]v_j¹u_i¹a^limju_iv_j modg_k‡1(G⁰);

we obtain

gf(a)Y^r

iˆ1

Y^s

jˆ1

v_j¹u_i¹a^limju_iv_jfg(a) modg_k‡1(G⁰):

It follows g ¹f ¹gf(a)a modg_k‡1(G⁰), which is the desired

result. p

PROOF OFTHEOREM1. Thus we suppose now thatGis (nilpotent of class c)-by-abelian, and sog_c‡1(G⁰) is trivial. Recall thatKis the kernel of the the homomorphismF:Aut_n(G)!Aut_n(G=G⁰) defined above. The groupG=G⁰ beingabelian, each normal automorphism of G=G⁰ is in fact a power au- tomorphism, that is, an automorphism fixingsetwise every subgroup of G=G⁰. Since the group of all power automorphisms of a group is always abelian [1], so is Aut_n(G=G⁰). The group Aut_n(G)=Kbeingisomorphic with a subgroup of Aut_n(G=G⁰), it is abelian, and so Aut_n(G)=K⁰is metabelian. It remains to see thatK⁰ is nilpotent, of class at mostc. For that, we notice thatK⁰stabilizes the normal series (of lengthc‡1)

1ˆg_c‡1(G⁰)/g_c(G⁰)/ g₂(G⁰)/g₁(G⁰)ˆG⁰/G:

Indeed, the induced action ofK⁰ on the factorG=G⁰is trivial since every element of K is an IA-automorphism. On the other factors, it is a con- sequence of the second part of the lemma above. By a well-known result of KaluzÏnin (see for instance [8, p. 9]), it follows thatK⁰is nilpotent of classc,

as required. p

PROOF OFTHEOREM2. First notice that in an abelian group which is either finite or infinite non-periodic, the group of all power automorph- isms is finite. That is trivial when the group is finite. In the second case, the group of power automorphisms has order 2, the only non-identity power automorphism beingthe inverse functionx7!x ¹(see for instance [1, Corollary 4.2.3] or [6, 13.4.3]). Therefore, comingback to the proof of Theorem 1 when G=G⁰ is either finite or infinite non-periodic, we can assert that Aut_n(G)=K is finite. Since K is (nilpotent of class c)-by- abelian, the result follows.

PROOF OF THEOREM 3. Let G be a supersoluble group. Since G is polycyclic, so is Autn(G) [4]. Thus Autn(G) is finitely generated. Let us prove now that Autn(G) is nilpotent-by-(finite and supersoluble). By a

result of Zappa [6, 5.4.8], there is a normal series 1ˆG_m/G_{m 1}/ /G₁/G₀ˆG …1†

in which each factor is cyclic of prime or infinite order. For any k2 f0;1;. . .;mg, we denote byG_k the set of all normal automorphisms of G which stabilize the series G_k/G_{k 1}/ /G₁/G₀ˆG (we put G₀ˆAut_n(G)). Clearly, G₀;G₁;. . .;G_m forms a decreasingsequence of normal subgroups of Aut_n(G). Usingonce again the result of KaluzÏnin [8, p. 9], we can assert thatGmis nilpotent (of class at mostm 1) since Gmstabilizes the series (1) above. It remains to prove that Autn(G)=Gm

is supersoluble and finite. For each integer k (with 0km 1), consider the homomorphism C_k:G_k!Aut(G_k=G_k‡1), where for any f 2G_k,C_k(f) is defined as the automorphism induced byf onG_k=G_k‡1. We observe that Aut(Gk=Gk‡1) is finite cyclic and that kerCkˆGk‡1. Consequently, the factor G_k=G_k‡1 is finite cyclic, since it is isomorphic to a subg roup of Aut(G_k=G_k‡1). It follows that

1ˆ…Gm=Gm†/…Gm 1=Gm†/ /…G1=Gm†/…G0=Gm† ˆ…Autn(G)=Gm† forms a normal series in which each factor is finite cyclic. Thus Aut_n(G)=G_m is supersoluble and finite and the proof is complete. p

REFERENCES

[1] C. D. H. COOPER,Power automorphisms of a group, Math. Z.,107(1968), pp.

335-356.

[2] G. ENDIMIONI, Pointwise inner automorphisms in a free nilpotent group, Quart. J. Math.,53(2002), pp. 397-402.

[3] G. ENDIMIONI,Normal automorphisms of a free metabelian nilpotent group, http://arxiv.org/abs/math.GR/0612347.

[4] S. FRANCIOSI and F. DE GIOVANNI, On automorphisms fixing normal subgroups of nilpotent groups, Boll. Un. Mat. Ital. B,7(1987), pp. 1161-1170.

[5] A. LUBOTZKY,Normal automorphisms of free groups, J. Algebra,63(1980), pp. 494-498.

[6] D. J. S. ROBINSON,A course in the theory of groups, Springer-Verlag, 1982.

[7] D. J. S. ROBINSON,Automorphisms fixing every subnormal subgroup of a finite group, Arch. Math.,64(1995), pp. 1-4.

[8] D. SEGAL,Polycyclic groups, Cambridge University Press, 1983.

Manoscritto pervenuto in redazione il 2 aprile 2007.