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CHARACTERIZATION OF THE AUTOMORPHISMS HAVING THE LIFTING PROPERTY IN THE CATEGORY
OF ABELIAN p -GROUPS
S. ABDELALIM and H. ESSANNOUNI Received 23 October 2002
Letpbe a prime. It is shown that an automorphismαof an abelianp-groupAlifts to any abelianp-group of whichAis a homomorphic image if and only ifα=πidA, withπan invertiblep-adic integer. It is also shown that ifAis a torsion group or torsion-freep-divisible group, then idAand−idAare the only automorphisms of Awhich possess the lifting property in the category of abelian groups.
2000 Mathematics Subject Classification: 20K30.
1. Introduction. Every inner automorphism of a groupGhas the property that it extends to an automorphism of any group containingG as subgroup.
Schupp [4] showed that this extension property characterizes inner automor- phisms in the category of groups. Pettet [3] gave an easier proof of Schupp’s result and proved at the same time that the inner automorphisms of a group Gare also characterized by the lifting property in the category of groups. In [1], we characterized the automorphisms of abelianp-groups having the ex- tension property in the category of abelianp-groups, as well as those having the extension property in the category of all abelian groups.
LetᏯbe a full subcategory of the category of abelian groups. An automor- phismαofA∈Ꮿhas the lifting property inᏯif, for allB∈Ꮿand any epimor- phisms:B→A, there existsα∈Aut(B)such thats◦α=α◦s, in other words, the diagram
B
α
s A
α
B s A
(1.1)
commutes. In this note, we show that an automorphismαof ap-groupA(with pbeing a prime number) has the lifting property in the category of abelianp- groups if and only ifα=πidA, withπ an invertiblep-adic number. We also determine the automorphisms of an abelian groupAhaving the lifting property in the category of all abelian groups, whenAis either torsion or p-divisible torsion-free. In both cases they are idAand−idA.
We will use the notation introduced in [2].
2. The lifting property in the category of thep-groups. Letpbe a prime number.
Lemma2.1. Letαbe an automorphism of a p-groupA having the lifting property in the category of abelianp-groups. IfCis subgroup ofAwithα(C)= C, then the restriction ofαtoCalso has the lifting property in the category of abelianp-groups.
Proof. Letµ:B→C→0 be an exact sequence. It follows from [2, page 108]
that we have a commutative diagram with exact rows:
0 Kerµ i B µ
σ
C
j
0
0 Kerµ λ F γ A 0,
(2.1)
whereiandj are the canonical injections. It is easy to show thatF is again a p-group, then there exists α ∈Aut(F ) such that γα =αγ. If we put, for anyb∈B, α(σ (b)) =σ (γ(b)), thenγ∈Aut(B)andµγ=α0µ, withα0the restriction ofαtoC.
Lemma 2.2. LetA be a torsion group and n∈ N∗. Then there exists an abelian groupBand an epimorphismµ:B→Asuch thatB[n]⊆Kerµ, where B[n]= {b∈B|nb=0}.
Proof. Fora∈A, we putBa= xa, whereo(xa)=o(a)andµa:Ba→A is defined byµa(xa)=a. If we putB=
a∈ABa andµ:B→A, whereµ(xa)= µa(xa), for alla∈A, thenµis an epimorphism andB[n]⊆Kerµ.
Theorem2.3. LetAbe an abelianp-group and an automorphismαofAhas the lifting property in the category of abelianp-groups if and only ifα=πidA, whereπis an invertiblep-adic number.
Proof. One implication is clear. Assume thatαhas the lifting property in the category of abelianp-groups. The proof of the fact thatα=πidAgoes in three steps.
Step1. We suppose thatAis reduced. Letx∈Abe such thatxis a direct summand ofA. We prove thatα(x)∈ x.
Putx
A=Aand letE(A)be the injective envelope ofA. We put A=
y∈E A
|pny∈A
, (2.2)
whereo(x)=pn. We consider the group B= x
A; the map s :B→A defined by
s(mx+y)=mx+pny, (2.3)
for all m∈Z and y ∈A, is an epimorphism. Therefore, there exists α ∈ Aut(B)such thatsα=αs. We can writeα(x) =kx+a, withk∈Zanda∈ A. Now
sα(x) =kx+pna=kx=αs(x)=α(x) (2.4) because pna =0, thus α(x)∈ x. Let B be a basic subgroup ofA, B =
n≥1Bn, and, for any n≥1, Bn=0 orBn is a direct sum of torsion cyclic groups of orderpn. We supposeBn≠0 forn≥1, soBn=
i∈Ixisuch that o(xi)=pn, for alli∈I, sinceBnis a direct summand ofA(see [2, page 138]).
Withmi∈Z,α(xi)=mixi. Let(i, j)∈I2withi≠j. We can writeA= xi Ai
withxj∈Ai. It is easy to see thatxi+xj
Ai=A, soα(xi+xj)=m(xi+xj), hencepn|(mi−mj). Then there iskn∈Zsuch thatα(b)=knb, for allb∈Bn. For(m, n)∈N2where 1≤m < n,Bm
Bnis a direct summand ofA[2, page 138] and it is easy to see thatpm|(kn−km).
Letπ be thep-adic number defined by(kn)n≥0(withk0=0 andkn=kn−1
ifBn=0). Thenα(b)=π b, for allb∈B. SinceAis reduced, it follows that α=πidA(see [2, page 145]).
Step2. We suppose thatAis divisible. Therefore,A=
i∈IAiwith Ai Z(p∞), for all i∈I (see [2, page 104]). We consider the direct product E=
n≥1xn, whereo(xn)=pn, for alln≥1. For alln≥1, leten∈Ebe defined by
fm
en
=
0 ifm < n,
pm−nxm ifm≥n, (2.5)
wherefm:E→ xmis the canonical projection. LetC be the following sub- group ofE:
C=
n≥1
xn +
en|n≥1
. (2.6)
It is easy to see thatC/(
n≥1xn)Z(p∞).
We choosei∈Iandai∈Ai. We want to show thatα(ai)∈Ai. Letj∈Iwith j≠i. We putA=
k∈I−{j}Akand we haveA=Aj
A. Letγ:C→Ajbe an epimorphism. If we suppose thatB=C
Aand considers:B→Awhich is defined bys(c+a)=γ(c)+a (c∈C,a∈A), thens is an epimorphism.
Therefore, there existsα∈Aut(B)such thatsα=αs.SinceAis a maximal divisible subgroup of B, α(a )=a. Sinceai∈A, thenα(a i)=α(ai)∈A. Thus for allj≠i,α(ai)∈
k≠jAk, and therefore,α(ai)∈Ai. Then there is ap-adic numberπisuch thatα(ai)=πiai, for allai∈Ai(see [2, page 181]).
For eachi∈I, we putAi= {yi,n|n≥1}withpyi,1=0 andpyi,n+1=yi,n, for alln≥1. Let(i, j)∈I2withi≠j. If we suppose thatzn=yi,n+yj,nand H= {zn|n≥1}, thenHZ(p∞)andAi
Aj=Ai
H. By the preceding
arguments, there exists ap-adic numberπsuch thatα(h)=π h,αh∈H. Then we deduce thatπi=πj=π.
Step3. We suppose thatAis an arbitrary abelian p-group. We can write A=C
D withC reduced andDdivisible. We can also suppose that C≠0 andD≠0. We haveα(D)=D, and the restrictionα1ofαtoDhas the lifting property in the category of p-groups, byLemma 2.1. Then there is ap-adic numberπsuch thatα(d)=π d, for alld∈D.
Letc0∈Cwitho(c0)=pn0. we define the maps:A→Aby
s(c+d)=c+pn0d, (2.7)
for(c, d)∈C×D. Thensis an epimorphism, and therefore, there existsα∈ Aut(A)such thatsα=αs. Putα(c 0)=c1+d1. Then
sα c0
=c1+pn0d1=c1=αs c0
=α c0
, (2.8)
and it follows thatα(c0)∈Candα(C)=C. We show thatα(c)=π c, for all c∈C. To this end, take
i∈Icias a basic subgroup ofC. We choosei∈I;
ciis a direct summand ofC. Putpni=o(ci)and
Ci=C. Letdi∈Dsuch thato(di)=pni. We have
A= ci+di
Ci
D. (2.9)
Then there exist a group G and an epimorphism η:G→Ci
D such that G[pni]⊆kerη, byLemma 2.2. We suppose thatB= ci+di
G, and we define µ:B→Gbyµ(m(ci+di)+g)=m(ci+di)+η(g). Thenµis an epimorphism.
Letα∈Aut(B)be such thatαµ=µα. We have αµ
ci+di
=α ci+di
=α ci
+π di. (2.10)
We put α(c i+di)=k(ci+di)+g0, then µα(c i+di)= k(ci+di) (because η(g0)=0). Thusα(ci)+π di=kci+kdi, soα(ci)=π ci, and therefore,α(c)= π c, for allc∈C, by [2, page 145].
3. The lifting property in the category of abelian groups. In this section, we show that, for a torsion orp-divisible torsion-free groupA(p is a prime number), idA and −idA are the only automorphisms ofA having the lifting property in the category of abelian groups.
Proposition3.1. LetAbe an abelian torsion group. Then an automorphism αofAhas the lifting property in the category of abelian groups if and only if α=ida orα= −ida.
Proof. One implication is obvious. Assume thatαhas the lifting property in the category of abelian groups and consider the exact sequence
E: 0 →Z →Q →Q/Z →0, (3.1)
then, by the Cartan-Eilenberg theorem (see [2, page 218]), the sequence 0=Hom(A,Q) →Hom(A,Q/Z) E∗→Ext(A,Z) →Ext(A,Q)=0 (3.2) is exact, whereE∗is the map associating toξ∈Hom(A,Q/Z)with the class extensionEξ.
LetE1: 0→Z→λ B →µ A→0 be an extension of ZbyA. Then there exists σ∈Aut(Z)such that the following diagram is commutative:
0 Z
σ
λ B µ
α
A
α
0
0 Z λ B µ A 0.
(3.3)
Ifσ =idZ, thenE1≡E1α, and ifσ = −idZ, thenE1≡E1(−α). Therefore, for allξ∈Hom(A,Q/Z),E∗(ξα−ξ)=0 orE∗(ξα+ξ)=0. Thusξ(α−id)=0 or ξ(α+id)=0, for allξ∈Hom(A,Q/Z).
From the fact thatQ/Zis divisible, it follows thatα=id orα= −id.
Proposition3.2. Letpbe a prime number andAap-divisible torsion-free group. Then an automorphismαofAhas the lifting property in the category of abelian groups if and only ifα=idaorα= −ida.
Proof. One implication is obvious. Suppose thatαhas the required lifting property, and consider the pure exact sequence
E: 0 →Z →Jp →Jp/Z →0, (3.4) whereJpis the additive group ofp-adic integers. By the theorem of Harrisson (see [2, page 231]), the sequence
Hom A, Jp
→Hom
A, Jp/Z E∗
→Pext(A,Z) →Pext A, Jp
(3.5)
is exact. Hom(A, jp)=0 becauseJpcontains no nonzerop-divisible subgroup and Pext(A, jp)=0 becauseJpis algebraically compact. ThusE∗is an isomor- phism, and, as in the proof ofProposition 3.1, we find thatα=id orα= −id.
References
[1] S. Abdelalim and H. Essannouni,Caractérisation des automorphismes d’un groupe abélien ayant la propriété de l’extension, Portugal. Math.59(2002), no. 3, 325–333 (French).
[2] L. Fuchs,Infinite Abelian Groups. Vol. I, Pure and Applied Mathematics, vol. 36, Academic Press, New York, 1970.
[3] M. R. Pettet,On inner automorphisms of finite groups, Proc. Amer. Math. Soc.106 (1989), no. 1, 87–90.
[4] P. E. Schupp,A characterization of inner automorphisms, Proc. Amer. Math. Soc.
101(1987), no. 2, 226–228.
S. Abdelalim: Department of Mathematics and Computer Science, Faculty of Sciences, Mohammed V University, B.P.1014 Rabat, Morocco
E-mail address:[email protected]
H. Essannouni: Department of Mathematics and Computer Science, Faculty of Sci- ences, Mohammed V University, B.P.1014 Rabat, Morocco
E-mail address:[email protected]
