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On Conditions for an Endomorphism to Be an Automorphism Algebra Colloquium 12 : 4 (2005) 709{714 Algebra Colloquium °c 2005 AMSS CAS & SUZHOU UNIV On Conditions for an Endomorphism to be an Automorphism Alireza Abdollahi Department of Mathematics, University of Isfahan Isfahan 81746-73441, Iran E-mail: [email protected] G¶erardEndimioni CMI - Universit¶ede Provence, UMR-CNRS 6632 39, rue F. Joliot-Curie, F-13453 Marseille Cedex 13, France E-mail: [email protected] Received 9 October 2003 Communicated by Jiping Zhang Abstract. If K is a set of automorphisms of a group G, an endomorphism θ : G ! G is said to be K-pointwise if for each element t 2 G, there exists an element ' 2 K such that θ(t) = '(t). This generalizes the notion of pointwise inner automorphism. We show that in some special cases, a K-pointwise endomorphism is necessarily an automorphism (it is not true in general). 2000 Mathematics Subject Classi¯cation: 20E36 Keywords: endomorphism, automorphism, soluble group 1 Introduction By de¯nition, a pointwise inner automorphism of a group G is an automorphism θ : G ! G such that t and θ(t) are conjugate for any t 2 G. This notion appears already in the famous book of Burnside [1, Note B, p. 463]. There are many examples of groups (¯nite or in¯nite) admitting a pointwise inner automorphism which is not inner (see, for instance, [2], [9], [7] or [8], where these groups are besides nilpotent). On the other hand, each pointwise inner automorphism is inner for example in a free group [4] or in a free nilpotent group [3]. In the same way, we de¯ne a pointwise inner endomorphism of G as an endo- morphism θ : G ! G such that t and θ(t) are conjugate for any t 2 G. Equivalently, an endomorphism θ of G is pointwise inner if and only if θ(tG) ⊆ tG for each t 2 G, where tG is the conjugacy class of t in G. Clearly, a pointwise inner endomorphism is always injective. But it is not nec- essarily surjective. Indeed, let S(N) be the ¯nitary symmetric group on the set of 710 A. Abdollahi, G. Endimioni natural integers. In other words, S(N) is the set of permutations of N having ¯nite support, with the operation of composition (the support of a permutation f is the set fn 2 N j f(n) 6= ng). Consider the mapping θ : S(N) ! S(N), where for each element f 2 S(N), the permutation θ(f) is de¯ned by θ(f)(0) = 0; and θ(f)(n) = 1 + f(n ¡ 1) when n > 0 (θ could be called the 1-right-shift). It is easy to verify that θ is an endomorphism of S(N). Moreover, for any f 2 S(N), we have θ(f) = (0 1 2 : : : k k+1) ± f ± (0 1 2 : : : k k+1)¡1; where the integer k is chosen such that the set f0; 1; 2; : : : ; kg contains the support of f. Therefore, θ is a pointwise inner endomorphism. But θ is not surjective since θ (S(N)) = ff 2 S(N) j f(0) = 0g 6= S(N): However, we shall see that in a soluble group, each pointwise inner endomorphism is surjective (and so is an automorphism). In fact, we shall prove a slightly stronger result. For that, we shall consider the notion of pointwise inner endomorphism as a particular case of a more general concept. 2 Results We denote by A(G) the set of automorphisms of a group G and by I(G) the set of inner automorphisms. Let K be a subset of A(G). An endomorphism θ : G ! G is said to be K-pointwise if for each element t 2 G, there exists an element ' 2 K such that θ(t) = '(t). When K is a subgroup, that means that for each t 2 G, we have θ(tK ) ⊆ tK , where tK denotes the orbit of t under the action of K. In particular, when K = I(G), the notions of \pointwise inner endomorphism" and \K-pointwise endomorphism" coincide. Obviously, any K-pointwise endomorphism is injective. We aim in this paper to show that under a suitable hypothesis on G and K, each K-pointwise endomorphism is an automorphism. For example, suppose that K is a ¯nite subgroup of A(G). K K A K-pointwise endomorphism θ induces an injection θt : t ! t for any t in G. K But since t is ¯nite, θt is bijective and so θ is an automorphism. It turns out that more generally, the result remains true when K is a ¯nite subset of A(G). Proposition 2.1. Let G be a group and K a ¯nite subset of A(G). Then each K-pointwise endomorphism θ : G ! G is an automorphism. In addition, the proof of this result will show that there exist a subgroup H · G of ¯nite index and an automorphism ' 2 K such that θ(x) = '(x) for all x 2 H. If jKj = 2, θ is actually in K but this is not always the case when jKj ¸ 3 (see the remark following the proof of Proposition 2.1 in the next section). Now recall that a normal automorphism ' of a group G is an automorphism such that '(H) = H for each normal subgroup H of G. We shall denote by An(G) the set of normal automorphisms of G. On Conditions for an Endomorphism to be an Automorphism 711 Proposition 2.2. In a soluble group G, each An(G)-pointwise endomorphism is an automorphism. Notice that the automorphism obtained in the conclusion of this proposition is obviously a normal automorphism. Since An(G) contains I(G), each pointwise inner endomorphism is also an An(G)- pointwise endomorphism. Thus, as an immediate consequence of Proposition 2.2, we can state: Corollary 2.3. In a soluble group, each pointwise inner endomorphism is an automorphism. As mentioned in the introduction, this automorphism is not necessarily inner. Proposition 2.2 cannot be extended to A(G)-pointwise endomorphisms. In- deed, consider for example the free abelian group of countably in¯nite rank. If fe0; e1; e2;::: g is a free basis, the endomorphism θ de¯ned by θ(ei) = ei+1 (i = 0; 1; 2;::: ) is A(G)-pointwise but is not an automorphism. However, we have: Proposition 2.4. Let G be a soluble group of ¯nite rank. Then each A(G)- pointwise endomorphism is an automorphism. Recall that a group is said to be of ¯nite rank (in the sense of PrÄufer)if there is a positive integer n such that every ¯nitely generated subgroup can be generated by n elements. We do not know whether in a ¯nitely generated soluble group G, each A(G)- pointwise endomorphism is an automorphism. Nevertheless, we can positively an- swer in a particular case. Proposition 2.5. In a 2-generated nilpotent-by-abelian group G, each A(G)- pointwise endomorphism is an automorphism. We end this section with a question: Question. Suppose that K = I(G) or K = A(G). Is it possible to ¯nd a ¯nitely generated group G and a K-pointwise endomorphism θ : G ! G which is not an automorphism? Obviously, a positive answer when K = I(G) would imply a positive answer when K = A(G). 3 Proofs In the following, we shall use the evident fact that any K-pointwise endomorphism is injective. The proof of the ¯rst lemma is straightforward and is omitted. Lemma 3.1. Let H be a normal subgroup of a group G. Let θ be an endomorphism of G such that θ(H) · H. Denote by θ (respectively, θ0) the endomorphism induced by θ in G=H (respectively, in H). If θ and θ0 are surjective, then so is θ. 712 A. Abdollahi, G. Endimioni Lemma 3.2. In a group G, consider an injective endomorphism ¿. Suppose that the subgroup H = fx 2 G j ¿(x) = xg is of ¯nite index in G. Then ¿ is an auto- morphism. Proof. We must prove that ¿ is surjective. Recall that the core HG of H in G is the T ¡1 normal subgroup of G de¯ned by HG = x2G x Hx. Since H is of ¯nite index in G, so is HG. Clearly, ¿ induces in the subgroup HG the identity automorphism, and in the quotient G=HG an injective endomorphism. But this quotient is ¯nite, so ¿ induces an automorphism in G=HG. Hence, it follows from Lemma 3.1 that ¿ is surjective. 2 Proof of Proposition 2.1. Suppose that K = f'1;:::;'ng. For each integer i = ¡1 1; : : : ; n, consider the subgroup fx 2 G j ¿i(x) = xg, where ¿i = 'i ± θ. Notice that ¿i is injective. Since θ is K-pointwise, we have G = H1 [ ¢ ¢ ¢ [ Hn. It follows then from a well-known result of B.H. Neumann [5] that among the subgroups H1;:::;Hn, at least one, say Hj, is of ¯nite index in G. By Lemma 3.2, ¿j is an automorphism; hence so is θ. 2 Remark. We keep here the notation used in the proof of Proposition 2.1. If n = 2, then G = H1 [ H2. It is well known and easy to see that such an equality implies G = H1 or G = H2. In other words, θ = '1 or θ = '2. On the other hand, if K contains three elements, a K-pointwise endomorphism is not always in K. For example, let G be the Klein 4-group. This group can be considered as a vector space of dimension 2 over the ¯nite ¯eld of order 2.
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