R.J. Apon

Aspects of automorphism towers

Thesis, December 9, 2016

Supervisor: prof.dr. H.W. Lenstra

Mathematical Institute, University of Leiden Contents

1 Introduction 3

2 Definitions, conventions and an essential lemma 4

3 Automorphism group and normalizer 5

4 Specific height 7

5 Automorphism tower of p-adic integers 10

6 Automorphism groups of finite products of abelian groups 14

7 Abelian automorphism towers for finitely generated abelian groups 16

8 Basic subgroups 18

9 Abelian automorphism groups 22

10 References 24

2 1 Introduction

−1 For a group G denote θG : G → Aut G, g 7→ (x 7→ gxg ). Whenever the group G is clear from the context, it will be omitted from the notation. Define G0 = G and for i ∈ Z≥1 i define Gi = Aut Gi−1. I use the notation Aut G for Gi. This sequence of groups combined with the maps θGi is called the automorphism tower of G. When introducing a new definition, new questions arise. I cannot answer all questions in this thesis, hence I focused on three aspects of the automorphism towers: the height, stable categories and the abelian automorphism towers.

Wielandt ([1], 1939) showed that for a finite group with a trivial center there exists m ∈ Z≥0 such that θGm is an isomorphism. It would be logical to give such smallest m a name. We say that a group G has height m if m is the smallest non-negative integer such that ∼ Gm = Gm+1. Note that in my definition of height I do not specify what the isomorphism between the groups is, because later on I will use the definition for abelian groups as well. The height of a finite group with a trivial center is finite, but can we reach all possible heights? The answer is yes and is shown by the following theorem.

Theorem 1.1. For all m ∈ Z≥0 there exists a finite group G with trivial center such that

G has height m and the map θGm is an isomorphism. This will be proven in chapter 4. One could also ask the question if we can find an infinite group with a trivial center such that its automorphism tower has infinite height. The contruction used for theorem 1.1 can be extended to get such a group. Since it takes little effort, I will show this in chapter 4 as well. Calculating automorphism towers can become quite difficult. For example the automor- phism tower of the group of 2-adic integers Z2 is already quite difficult after 6 steps (calcu- lating the tower for Z3 is a fun exercise). This is the reason that I don’t work out specific examples. It would be nice if we could formulate some kind of theorem that shows that if G has some property then Aut G has that as well. These kinds of theorems show that every group in the automorphism tower of G has a certain property. This inspired the following definition. For a category C denote C0 the collection of objects and C1 the collection of morphisms. Let C be a full subcategory of the category of groups. We call C stable if for ∼ all C ∈ C0 there exists a D ∈ C0 such that Aut C = D. The following stable category is actually used in the proof of theorem 1.1. A group G is called semi-simple if G is isomorphic to a direct sum of simple groups and is called perfect if G = [G, G].

Theorem 1.2. Let S be a perfect semi-simple group. Then the category CS with (CS)0 = {G ⊂ Aut S|θS(S) ⊂ G} is a stable category. I will prove this theorem in chapter 3. As said before, I calculated the automorphism tower of Z2 for a few steps. I could not find out whether this tower has finite height or not, but I could deduce a pattern which led to the following result.

Theorem 1.3. Let p be a prime. Then the category (Cp)0 = {G|∃f : G  Zp : # ker(f) < ∞} is a stable category. I will prove this in chapter 5. I will also show that the existence of a surjective group homomorphism G  Zp with a finite kernel is equivalent to two other statements. Let A be an abelian group. If all groups in the automorphism tower of A are abelian,

3 then A has an abelian automorphism tower. I have tried to classify all abelian groups that have an abelian automorphism tower. For finitely generated abelian groups I have found a classification. For infinitely generated abelian groups I do not have a classification, but I do have a result that eliminates a lot of groups. Theorem 1.4. Let A be a finitely generated abelian group. Then A has an abelian auto- morphism tower iff A is cyclic and |A| is contained in one of the following sets: m m a) {3 , 2 · 3 |m ∈ Z≥1} m m m b) {2 · 3 + 1, 2(2 · 3 + 1)|m ∈ Z≥1, 2 · 3 + 1 is prime} c) {1, 2, 4, 5, 10, 11, 22, 23, 46, 47, 94} d) {∞} m This theorem will be proven in chapter 7. Note that Z/3 Z has height m + 1 for m ≥ 1. This shows that for each m ∈ Z≥0 we can find a group that has an abelian automorphism tower of height m. From theorem 1.4 we can deduce that if we want to find a group with an abelian automorphism tower of infinite height, we need to look at infinitely generated abelian groups. Theorem 1.5. Let A be an abelian group such that Auti A is abelian for i = 0, ..., 4 and ∼ −1 Aut A is infinite. Then Aut A = Z/2Z ⊕ C where C is a Z[2 ]-module. I will prove this in chapter 9, which uses p-basic subgroups. I do not assume that the reader knows this, hence chapter 8 covers this topic. I have not been able to show that for each m ∈ Z≥0 there exists an infinitely generated abelian group such that its abelian automorphism tower has height m (finding a non-trivial group for m = 0 is a fun exercise), which is considerably harder than in the finitely generated case. I did not show either there exists a group with an abelian automorphism tower of infinite height.

2 Definitions, conventions and an essential lemma

For a set X denote Sym X as the symmetric group acting on X. For n ∈ Z≥0 denote n as the set {0, 1, ..., n − 1}. X Q (X) L For a group G and a set X I will denote G := x∈X G and G := x∈X G. I will denote Gx as the x-th coordinate axis. For a non-trivial ring R with a multiplicative unit, denote R∗ as the group of invertible elements of R. Definition 2.1. Let G be a group. The exponent of G is the least common multiple of the orders of all elements in G. If there is no least common multiple, the exponent is infinite. Definition 2.2. Let G and H be groups and Ω an H-set. Define ψ : H → Aut(GΩ), h 7→ Ω ((gw)w∈Ω 7→ (gh−1w)w∈Ω). Define the wreath product G oΩ H := G oψ H. I will often leave out Ω, since it should be obvious from the context which set is used in the definition. An abelian group will be written additively and 0 is the identity element. The only exceptions to this rule are automorphism groups. I will not denote automorphism groups additively even if they are abelian, since this would be more confusing. I will also denote idG as the identity element of Aut G and ◦ as the symbol for composition. Groups which

4 are not specified to be abelian will be written multiplicatively and 1 is the identity element. Whenever I write 1 or 0 as a group, I mean the trivial group.

f g Lemma 2.3. Let 1 → A −→ B −→ C → 1 be a short exact sequence of groups and s : C → B a homomorphism such g ◦ s = idC . Then B is isomorphic to A oψ C where ψ : C → Aut A, c 7→ (x 7→ g(c)xg(c)−1).

Proof. See lemma 2.1 of [2], which is a more general version. 

3 Automorphism group and normalizer

In this chapter I will prove two theorems that show a connection between taking automor- phism groups and taking normalizers. Note that theorem 3.2 directly implies theorem 1.2. Theorem 3.1 will be needed for the next chapter. Theorem 3.1. Let T be a non-abelian simple group, X a set and H ⊂ Sym X a subgroup. ∼ Then Aut((Aut T ) o H) = (Aut T ) o NSym X (H). Theorem 3.2. Let S be a perfect semi-simple group. Let G ⊂ Aut S be a subgroup with −1 θS(S) ⊂ G. Then ψ :NAut S(G) → Aut G, x 7→ (g 7→ xgx ) is an isomorphism. Before proving these theorems, I will show some easy results and recall some defini- tions. L Lemma 3.3. Let I be a set, Si simple groups for i ∈ I and define S = i∈I Si. Then S is perfect iff for all i ∈ I holds that Si is non-abelian. L Proof. Follows from [S, S] = i∈I [Si,Si].  Whenever I denote S a perfect semi-simple group, I will always use the letters I for the set and Si for the simple non-abelian groups. Since these letters will stay fixed throughout this chapter, I will not define them any more. −1 Theorem 3.4. Let σ ∈ Aut G and g ∈ G. Then holds: θG(σ(g)) = σ ◦ θG(g) ◦ σ . This theorem is trivial, writing out the definitions will just give the equality. However I state it as a theorem since I will be using it a lot.

Lemma 3.5. Let G be a group with a trivial center. Then CAut G(θG(G)) is trivial and Aut G has a trivial center.

Proof. We know ker(θG) = Z(G) = 1, so θG is injective. Let σ ∈ Aut G such that σ commutes with θG(g) for all g ∈ G. By theorem 3.4 and the commuting property, we get θG(g) = θG(σ(g)). This implies σ = idG, hence CAut G(θG(G)) = 1. Note that the inclusion Z(Aut G) ⊂ CAut G(θG(G)) always holds, hence Aut G has a trivial center.  Lemma 3.6. Let S be a perfect semi-simple group. Then any non-trivial normal subgroup N C S contains Si for some i ∈ I. Also, the set of minimal normal subgroups of S equals {Si|i ∈ I}. Proof. Let N be a non-trivial normal subgroup of S. Suppose that for all i ∈ I we have N ∩ Si = 1. Now for all i ∈ I we have [N,Si] ⊂ N ∩ Si, so N and Si centralize each other. Since S is the direct sum of these Si, we see that N ⊂ Z(S). Since all Si are non-abelian and simple, we must have Z(S) = 1. Since N is non-trivial, we get a contradiction. So there exists i ∈ I such that [N,Si] = Si ⊂ N.

5 Let i ∈ I and N C S be non-trivial with N ⊂ Si. By the above, N contains some Sj for j ∈ J. Clearly if i 6= j we have Si ∩ Sj = 1, hence we must have i = j. This shows N = Si, hence Si is a minimal normal subgroup.

Let N be a minimal normal subgroup of S. Let i ∈ I such that Si ⊂ N. Now Si is normal in S and N was minimal, hence Si = N. So the set of minimal normal subgroups is {Si|i ∈ I}.  Definition 3.7. Let G be a group. The socle of G is the subgroup generated by all minimal normal subgroups of G. It is denoted by Soc (G). Definition 3.8. Let H ⊂ G be groups. Then H is called a characteristic subgroup if for all σ ∈ Aut G we have σ(H) = H. Lemma 3.9. Let G be a group. Then Soc (G) is a characteristic subgroup of G.

Proof. This is left as an exercise for the reader.  The proof of theorem 3.1 uses theorem 3.2, so I will prove 3.2 first.

Proof of 3.2. In the rest of the proof I will only use the group S for θS, therefore I will shorten notation to θ. I will first show that θ(S) is a characteristic subgroup of G, which I will do by showing that θ(S) is actually the socle of G. I will first show Soc (G) ⊂ θ(S). Let N be a minimal normal subgroup of G. Note that θ(S) is the inner automorphism group, hence it is normal in Aut S. Therefore it is also normal in G. Now consider θ(S)∩N. It is normal in G, but also contained in N. Since N is a minimal normal subgroup, the intersection is trivial or equal to N. Suppose it is trivial. We have [N, θ(S)] ⊂ θ(S) ∩ N, so N and θ(S) centralize each other. But CAut S(θ(S)) is trivial by lemma 3.5. Any minimal normal subgroup must be non-trivial by definition, so we have a contradiction. Hence θ(S) ∩ N = N, or equivalently N ⊂ θ(S). To make the proof of θ(S) ⊂ Soc (G) easier, I will introduce a new definition. Definition 3.10. Let H ⊂ G be groups. Now define the normal closure (or also called conjugate closure) of H in G, denoted HG, as the smallest normal subgroup in G that contains H. In symbols we have HG = T N. H⊂NCG

To show θ(S) ⊂ Soc G, it is enough to prove that for all i ∈ I we have θ(Si) ⊂ Soc G. I will do this by constructing a minimal normal subgroup of G, that contains θ(Si). In G fact, this minimal normal subgroup is exactly θ(Si) . To show this, it is useful to know HG = hgHg−1|g ∈ Gi. This is left as an exercise for the reader. G Let i ∈ I and consider θ(Si) . Note that any automorphism of S must send minimal normal subgroups of S to minimal normal subgroups of S. Using theorem 3.4 and lemma −1 3.6 we get that for all σ ∈ Aut S we have σθ(Si)σ = θ(σ(Si)) = θ(Sj) for some j ∈ I. G Hence θ(Si) = hθ(Sj)|j ∈ Ji for some J ⊂ I. Take such a J. G Now all I need to show is that θ(Si) is actually a minimal normal subgroup of G. Suppose G −1 M is a non-trivial normal subgroup of G with M ⊂ θ(Si) . Note that θ (M) is also normal G in S, hence by lemma 3.6 it must contain some Sk for k ∈ I. In particular θ(Sk) ⊂ θ(Si) . Note that holds: θ(hSj|j ∈ Ji) = hθ(Sj)|j ∈ Ji.

Since θ is injective, we must have k ∈ J. Hence there exists g ∈ G such that θ(Sk) = −1 G G G G G gθ(Si)g . Now also θ(Si) ⊂ θ(Sk) , hence θ(Sk) = θ(Si) . From θ(Sk) ⊂ M ⊂ θ(Si) G follows M = θ(Si) .

6 So now I have proven θ(S) = Soc G, which is a characteristic subgroup of G by lemma 3.9. Now I will prove that ψ is an isomorphism. Clearly the kernel of ψ is exactly the centralizer of G in NAut S(G). But every element that centralizes G, also centralizes θ(S). However, from lemma 3.5, we know CAut S(θ(S)) = 1, hence the kernel must be trivial. This proves injectivity. Let σ ∈ Aut G. I will show σ ∈ Im(ψ). Since θ(S) is a characteristic subgroup, we have that σ|θ(S) is an automorphism. Take τ ∈ Aut S such that for all s ∈ S we have σ(θ(s)) = θ(τ(s)). Define γ : G → Aut S, α 7→ τ ◦ α ◦ τ −1. By theorem 3.4, we see that σ and γ are the same map on θ(S). Let h ∈ θ(S) and g ∈ G. Since θ(S) is normal in G we have: ghg−1 = γσ−1(ghg−1) = γσ−1(g)hγσ−1(g−1) ⇒ γσ−1(g−1)gh = hγσ−1(g−1)g −1 −1 ⇒ γσ (g )g ∈ CG(θ(S)).

From lemma 3.5 we know CG(θ(S)) = 1, hence σ(g) = γ(g). This shows that the image of γ is G and therefore we have τ ∈ NAut S(G) and ψ(τ) = σ. Since ψ is surjective and injective, it is an isomorphism.  (X) Proof of 3.1. Define S = T and denote Tx as the x-th coordinate axis. Note that T is non-abelian and simple, hence S is a perfect semi-simple group. I will first show that Aut T o Sym X is isomorphic to Aut S. We have a homomorphism f : (Aut T )X → Aut S, which is applying an automorphism of T on each coordinate of S. From lemma 3.6 follows:

∀α ∈ Aut S, x ∈ X : ∃y ∈ X : α(Tx) = Ty. From this we see that α induces a bijection on X. This way we get a homomorphism g : Aut S → Sym X. We have the following short exact sequence:

f g 0 (Aut T )X Aut S Sym X 0.

We can create a map s : Sym X → Aut S, which sends σ ∈ Sym X to the automorphism that permutes the coordinates of S using σ. Clearly gs = idSym X . Now using lemma 2.3, ∼ we have Aut S = (Aut T ) oψ Sym X. Note that ψ from lemma 2.3 is the same ψ as in the definition of the wreath product, hence we have Aut S =∼ (Aut T ) o Sym X. Denote the isomorphism Φ : (Aut T ) o Sym X → Aut S. Let H ⊂ Sym X and consider G = Φ((Aut T )oH). Note that we have θ(S) ⊂ Φ((Aut T )o1), ∼ hence θ(S) ⊂ G. Now G satisfies the requirements for theorem 3.2, so NAut S(G) = Aut G. ∼ Since Φ is an isomorphism, we have NAut S(G) = N(Aut T )oSym X ((Aut T ) o H). Note that the subgroup (Aut T ) o 1 is normal in (Aut T ) o Sym X and the quotient is isomorphic to Sym X. From the third isomorphism theorem we see that any subgroup of (Aut T )oSym X that also contains (Aut T ) o 1 must be of the form (Aut T ) o K for some subgroup K ⊂ Sym X. Also, the bijection from the third isomorphism theorem preserves conjugation. Thus N(Aut T )oSym X ((Aut T ) o H) is exactly (Aut T ) o NSym X (H). 

4 Specific height

In this chapter I will shorten my notation of taking normalizers. I will not denote the m group in which I take the normalizer, since this group is always Sym 2 for some m ∈ Z≥0. Note that A5 is a finite simple non-abelian group, hence theorem 4.1 implies 1.1.

7 m Theorem 4.1. Let m ∈ Z≥0. Then there exists a subgroup H ⊂ Sym 2 such that for each simple non-abelian group T the height of (Aut T ) o H is m, Z((Aut T ) o H) = 1 and θAutm((Aut T )oH) is an isomorphism.

Theorem 4.2. There exists a subgroup H ⊂ Sym Z≥0 such that for each simple non-abelian group T the height of (Aut T ) o H is infinite and Z((Aut T ) o H) = 1. m Theorem 4.3. For all m ∈ Z≥0 there exists a chain of subgroups H0 ⊂ ... ⊂ Hm ⊂ Sym 2 which satisfy the following statements:

1. N(Hi) = Hi+1 for 0 ≤ i < m, 2. N(Hm) = Hm, m 3. Hi acts transitively on 2 iff i = m, 4. |Hi+1 : Hi| = 2 for 0 ≤ i < m, 5. H0 stabilizes 0, m 6. Hm is a 2-Sylow subgroup of Sym 2 , m 7. H0 is a 2-Sylow subgroup of Sym (2 \{0}). To use theorem 4.3 and 3.1 for proving 4.1 conditions 3-7 are not needed. However, proving conditions 1 and 2 is easier if I show 3-5 first. Conditions 6 and 7 give a good idea of what the groups in the chains are. Before going into the proofs, I want to state a very important convention. I will be taking products of subgroups of symmetric groups, so I will define how to interpret this m m m+1 product. Let m ∈ Z≥0. Define X1 = {0, 1, ..., 2 − 1} and X2 = {2 , ..., 2 − 1}. m+1 m ∼ Note: X1 ∪ X2 = 2 . Since |X1| = |X2| = 2 , we have Sym X1 = Sym X2. So if we have subgroups G, H ⊂ Sym 2m, we can see G × H as a subgroup of Sym 2m+1 where for (g, h) ∈ G × H, x ∈ X1 and y ∈ X2 we have: (g, h)x = gx and (g, h)y = hy. The notation X1 and X2 will be used throughout this chapter.

Proof of 4.3. For m = 0, we can take H0 = Sym 1. This chain satisfies the needed m conditions trivially. Now suppose m ≥ 1 and we have a chain H0 ⊂ ... ⊂ Hm ⊂ Sym 2 with all the above properties. I want to create a chain which has m + 1 inclusions. Define G0 = H0 × Hm and define Gi+1 = N(Gi). I claim that the chain G0 ⊂ ... ⊂ Gm+1 ⊂ Sym 2m+1 satisfies the needed conditions. Conditions 1 and 5 trivally hold.

First I will show N(Hi × Hm) = Hi+1 × Hm for 0 ≤ i < m. The following fact about orbits is very useful: if Y ⊂ X is an orbit of H ⊂ Sym X, then for each α ∈ Sym X the set α(Y ) is an orbit of αHα−1. In particular, N(H) permutes the orbits of H. Let α ∈ N(Hi × Hm). Since Hm is transitive, we see that X2 is an orbit of Hi × Hm. Using the fact above, we see that α(X2) is an orbit of Hi × Hm. However, Hi does not act m m transitively on 2 so Hi × Hm has exactly one orbit of size 2 , which is X2. Therefore m+1 α(X2) = X2. Since X1 is the complement of X2 in 2 , we also have α(X1) = X1. We see that α must be contained in Sym X1 × Sym X2. If we write α = (g, h), we have −1 −1 −1 Hi × Hm = α (Hi × Hm)α = g Hig × h Hmh. So α ∈ N(Hi) × N(Hm). But by induction we know that this equals Hi+1 × Hm. So we know Gi = Hi × Hm for i ≤ m.

To show that the other conditions hold it is useful to give an explicit form of Gm+1. Denote Q2m−1 m m+1 γ = i=0 (i i + 2 ) ∈ Sym 2 , which is an element that switches X1 and X2. Note that γ normalizes Gm, hence γ ∈ Gm+1. I will show Gm+1 = Gm ∪ γGm. This proves that Gm+1 is isomorphic to Hm o Sym 2.

Let α ∈ Gm+1. Again, α(X2) is an orbit of Gm, but now we have two possibilities: α(X2) can be either X1 or X2. Suppose α(X2) = X2. Then α ∈ Sym X1 × Sym X2, and using the same argument as before we have α ∈ Gm. Suppose α(X2) = X1. Define β = γα. So

8 now β satisfies β(X1) = X1 and β(X2) = X2. Again using the same argument as before, 2 we see that β ∈ Gm. Since γ = 1, we have α = γβ.

From the induction hypothesis it is clear that |Gi+1 : Gi| = 2 for i < m. From the above, it is clear that this also holds for i = m. Therefore condition 4 is proven. Also Gm+1 acts m+1 transitively on 2 and Gi for i ≤ m does not, which proves condition 3. I will now show condition 2. Let α ∈ N(Gm+1). Since Gm+1 is transitive, we can take δ ∈ Gm+1 such that α(0) = δ(0). Define β = δ−1α, which stabilizes 0.

For any set X and subgroup K ⊂ Sym X denote (K)x as the subgroup of all elements which stabilize x ∈ X. The following holds:

m+1 β ∈ (Sym 2 )0 ∩ N(Gm+1) m+1 ⊂ N((Sym 2 )0) ∩ N(Gm+1) m+1 ⊂ N((Sym 2 )0 ∩ Gm+1)

= N((Gm+1)0).

All elements of G0 stabilize 0, therefore G0 ⊂ (Gm+1)0. From condition 4 follows |Gm+1 : m+1 G0| = 2 . Since Gm+1 is transitive, we see that Gm+1/(Gm+1)0 is in bijective cor- m+1 m+1 respondence to 2 . Hence the index of (Gm+1)0 is 2 . Therefore G0 = (Gm+1)0. This shows that β is contained in G1. Hence β ∈ Gm+1 and also α ∈ Gm+1. This shows N(Gm+1) = Gm+1.

Conditions 6 and 7 are left as an exercise for the reader. 

Proof of 4.1. Let m ∈ Z≥0 and T a simple non-abelian group. Take a chain of subgroups m H0 ⊂ ... ⊂ Hm ⊂ Sym 2 as in theorem 4.3. I will show that (Aut T ) o H0 satisfies the theorem. i From 3.1, 4.3.1 and 4.3.2, it follows that for i ≤ m holds Aut ((Aut T ) o H0) = (Aut T ) o Hi i and for i ≥ m holds Aut ((Aut T ) o H0) = (Aut T ) o Hm. Hence (Aut T ) o H0 has height at ∼ most m. Suppose (Aut T )oHi = (Aut T )oHj for some i and j. One can extend theorem 3.2, with an almost identical proof, to get the following theorem: let S be a perfect semi-simple group. Let G1,G2 ⊂ Aut S be subgroups with θS(S) ⊂ G1,G2. Then the following map is a bijection:

−1 ψ : {σ ∈ Aut S|σG1σ = G2} → {f ∈ Hom(G1,G2)|f is an isomorphism}, σ 7→ (x 7→ σxσ−1).

(2m) Denote S = T . For all k we have θ(S) ⊂ (Aut T ) o Hk ⊂ Aut S, hence we can use the extended version. Since (Aut T ) o Hi and (Aut T ) o Hj are isomorphic, we see that they are conjugate subgroups of Aut S. Therefore Hi and Hj are also conjugate. From cardinality follows i = j. From theorem 3.2 follows Z((Aut T ) o H0) ⊂ ker ψ = 1. Hence (Aut T ) o H0 has a trivial center. Note that θ(Aut T )oHm is an isomorphism, since this is exactly ψ from theorem 3.2. 

Proof of 4.2. For m ≥ 0 and 0 ≤ i ≤ m denote Gi,m as the i-th subgroup in the normalizer Q chain of length m as in 4.3. For k ∈ Z≥0 define Γk = m≥k Gm,m. For k ∈ Z≥0 denote Hk = Gk,k × Γk as a subgroup of Sym Z≥0.

Claim 4.4. Let i ∈ Z≥0. Then N(Hi) = Hi+1.

Claim 4.5. The center of (Aut T ) o H0 is trivial.

9 ∼ Claim 4.6. We have Aut T o Hi = Aut T o Hj iff i = j. The proofs are very similar to 4.1 and 4.3. The third claim also uses that conjugate n subgroups have the same multiset of orbit sizes. Note that Hi has orbits of size 2 iff n ≥ i. From the claims it should be clear that (Aut T ) o H0 satisfies the theorem. 

5 Automorphism tower of p-adic integers

Theorem 1.3 is a direct consequence of theorem 5.2. Theorem 5.1. Let p be a prime and G a group. Then the following are equivalent: ∼ (1) G = B oψ Zp for some finite group B and some homomorphism ψ : Zp → Aut B, (2) there exists a surjective homomorphism G → Zp with a finite kernel, (3) G is isomorphic to a subgroup of finite index of C × Zp for some finite group C. Theorem 5.2. If G satisfies (1)-(3) from theorem 5.2, then so does Aut G.

Lemma 5.3. Let G be a group with |[G, G]| < ∞. Then there exists m ∈ Z>0 such that for all g ∈ G holds gm ∈ Z(G). Proof. I use the notation [a, b] = aba−1b−1. Note #{[a, b]|a, b ∈ G} < ∞. Denote Gg = {hgh−1|h ∈ G}. Let g ∈ G, then we have:

#Gg = #Gg · g−1 = #{[x, g]: x ∈ G} ≤ #[G, G].

For convenience, denote n = #[G, G]. Consider the following exact sequence:

Q G 1 Z(G) G x∈G Sym x.

Note that, for all x, G acts on Gx by conjugation, which gives us a map G → Sym Gx. This way we get the map in the exact sequence. Now we know #Gx ≤ n, therefore we Q G see that the exponent of x∈G Sym x divides n!. Now it is clear that G/Z(G) has finite Q G exponent since x∈G Sym x has finite exponent. 

In the proofs of the theorems 5.1 and 5.2, I need quite a lot of properties of Zp. The focus in this thesis lies on understanding automorphism towers, not on working out details of properties of Zp. I will state the needed properties of Zp as lemmas. Readers familiar with Zp will most likely know them, other readers can use them as an exercise. ∼ ∗ Lemma 5.4. Let p be prime. We have Aut Zp = Zp.

Lemma 5.5. Let p be a prime and H ⊆ Zp be a subgroup of finite index. Then H is of m the form p Zp with m ∈ Z≥0, which is isomorphic to Zp. n ∼ n Lemma 5.6. Let p be a prime and n ∈ Z≥0. Then Zp/p Zp = Z/p Z. ∗ Lemma 5.7. Let p be a prime. Then Zp consists of exactly all elements of Zp that are ∗ ∼ ∗ ∼ not divisible by p. Also Z2 = Z2 ⊕ Z/2Z and for p > 2 we have Zp = Zp ⊕ Z/(p − 1)Z. m Lemma 5.8. Let p be a prime and m ∈ Z≥0. Then p Zp is a characteristic subgroup of Zp. The following lemma is somewhat trivial, but will be used throughout the proofs. Lemma 5.9. Let A ⊂ B and C be groups with |B : A| < ∞ and f : B → C a surjective group homomorphism. Then |C : f(A)| < ∞.

10 Proof. Left as an exercise for the reader.  0 Proof of theorem 5.1. (1) ⇒ (3). Define C = B oψ0 Aut B where ψ = idAut B. Clearly the map f : G → C × Zp, (b, x) 7→ (b, ϕ(x), x) is an injective homomorphism. Note that since B is finite, Aut B is finite. Therefore C is finite. Note that C × 0 is a complete set of representatives of all cosets of f(G) in C × Zp. So f(G) has finite index in C × Zp.

(3) ⇒ (2). Let π : C × Zp → Zp be the projection map. Since G is a subgroup of C × Zp of finite index, it follows from lemma 5.9 that π(G) is also of finite index in Zp. But by lemma 5.5 we see that π(G) is isomorphic to Zp. So there exists a surjective map from G ∼ to Zp. Note that ker π = C, which is finite.

(2) ⇒ (1). Let ϕ : G → Zp be a surjective map. I will be taking restrictions of the domain and codomain of ϕ very often. To avoid unpractical notation, I will only use the restriction of the domain in the notation. It should be clear from the context what the codomain should be. We have the following short exact sequence:

ϕ 1 ker(ϕ) G Zp 0.

I want to find a subgroup of G that is isomorphic to Zp, where a restriction of ϕ is this isomorphism. If we can do this, we can use lemma 2.3 to get the needed result. Note that G/ ker(ϕ) is abelian, hence [G, G] ⊂ ker(ϕ). The kernel of ϕ is finite, hence |[G, G]| < ∞. m From lemma 5.3, we can pick m ∈ Z>0 such that for all g ∈ G holds g ∈ Z(G). Now define H = h{gm|g ∈ G}i, which is a subgroup of Z(G). We get the following short exact sequence: ϕ|H 1 ker(ϕ) ∩ H H mZp 0. Denote by k the exponent of ker(ϕ) ∩ H and Hk = {hk|h ∈ H}. Note that Hk is a group k since H is abelian. We see that ϕ|Hk : H → mkZp is surjective and I will show that ϕ|Hk k k k is also injective. Let h ∈ H and suppose ϕ|Hk (h ) = 0. Note that Zp is torsion-free, therefore ϕ|Hk (h) = 0. We can conclude h ∈ ker(ϕ) ∩ H. Since k was exactly the exponent k k ∼ of this group, we have h = 0. Thus ϕ|Hk is injective. This shows H = mkZp, where a restriction of ϕ is this isomorphism. n Let γ ∈ G such that ϕ(γ) = 1. Write mk = p · l with p - l and n ∈ Z≥0. For convenience, denote ϕ| instead of ϕ|hγli·Hk . From lemma 5.7 follows that lZp is equal to Zp and from ∼ n lemma 5.6 we can conclude Zp/mkZp = Z/p Z. We have the following diagram:

0 Hk hγli · Hk hγlHki 0

ϕ|Hk ϕ| f π n 0 mkZp Zp Z/p Z 0

Note hγlHki ⊂ G/Hk. Define f by lifting an element to hγli · Hk and then applying π ◦ ϕ|. If a, b ∈ hγli · Hk are two different lifts, then a − b ∈ Hk. Using exactness and that the left square commutes we have π(ϕ|(a)−ϕ|(b)) = 0. This shows that f is well-defined. Note that l l k l n γ is a lift of the coset γ H . Also ϕ|(γ ) = l is a generator of Z/p Z because p - l, hence f is surjective. Also note that hγlHki has at most pn elements since γpn·l = γmk ∈ Hk. n n Since f is surjective and Z/p Z has p elements, it must be an isomorphism. The short five lemma states that if the diagram commutes, the two rows are short exact sequences and the maps ϕ|Hk and f are isomorphisms, then so is ϕ|. All these conditions −1 are true, hence ϕ| is also an isomorphism. Define g : Zp → G as the map ϕ| with a

11 bigger codomain. We have gϕ = idZp , thus from lemma 2.3 follows that G is isomorphic to ker(ϕ) o Zp.  In the next proof I use a little bit of group cohomology. The following two definitions will cover everything I need. Definition 5.10. Let G be a group, M an abelian group and ψ : G × M → M a map. We say that M is a left G-module if ψ is a left group action from G on M such that for all g ∈ G, a, b ∈ M we have g(a + b) = ga + gb, where ga denotes ψ(g, a). Definition 5.11. Let G be a group and M a left G-module. A crossed homomorphism or 1-cocycle is a map f : G → M such that for all g, h ∈ G holds f(gh) = f(g) + gf(h). The abelian group of crossed homomorphisms is denoted Z1(G, M). ∼ Proof of 5.2. Since G satisfies (1), we can take B such that G = B oψ Zp. It is enough to show that there exists a surjective homomorphism Aut(B oψ Zp) → Zp with finite kernel. Note that B is a characteristic subgroup of G, since B is the set of torsion elements of G, ∼ and G/B = Zp. Let σ ∈ Aut G. Note that σ : G/B → G/B, g 7→ σ(g) is a well-defined automorphism. This shows that we have a map f1 : Aut G → Aut Zp. I will now show that the image of f1 has finite index in Aut Zp. I will do this by showing that the subgroup H := {α ∈ Aut Zp : ψα = ψ} has finite index in Aut Zp and is contained in Im(f1). n From lemma 5.8 we have the canonical map γn : Aut Zp → Aut(Zp/p Zp). Write |Aut B| = m p · l with p - l for some m ∈ Z≥0. Let α ∈ ker(γm). We have that the following diagram commutes, where ψ is defined such that the right triangle commutes:

α ψ Zp Zp Aut B π π ψ m id m Zp/p Zp Zp/p Zp

From this follows ker(γm) ⊂ H. Clearly ker(γm) has finite index in Aut Zp, hence H has finite index in Aut Zp as well. For α ∈ H we can define gα : G → G, (b, x) 7→ (b, α(x)). From ψα = ψ follows that gα is an homomorphism. Since α is an automorphism, gα is as well. We can now conclude H ⊂ Im(f1), which shows that Im(f1) has finite index in Aut Zp.

I will now show that the kernel of f1 is finite. Define γ : ker(f1) → Aut B, σ 7→ σ|B. We have the following exact sequence:

γ 0 ker(γ) ker(f1) Aut B.

Note that for any exact sequence of groups G1 → G2 → G3 holds that if both G1 and G3 are finite, then so is G2. Hence it is enough to show that ker(γ) is finite. Let σ ∈ ker(γ), so we have σ|B = idB. Since ker(γ) ⊂ ker(f1), we have that by definition of f1 the following diagram commutes: π 0 B G Zp 0

id σ id π 0 B G Zp 0. ∼ −1 Note that the rows are exact since G/B = Zp. Define α : G → B, x 7→ σ(x)x . The map

12 α is not necessarily a group homomorhism. It is well-defined, since we have:

σ(x)x−1 ∈ B ⇔ π(σ(x)x−1) = 0 ⇔ π(σ(x)) = π(x).

The last statement is true since the diagram above commutes. Clearly σ(x) = α(x)x, so if I can show that there are at most finitely many α we are done. Let x ∈ G and y ∈ B. Since σ is the identity on B, we have:

α(xy) = σ(x)σ(y)y−1x−1 = σ(x)x−1 = α(x).

Note yx ∈ Bx = xB since B is normal in G. Therefore α(yx) = α(x). From this follows:

α(x) = α(yx) = σ(y)α(x)y−1 = yα(x)y−1.

This proves α(x) ∈ Z(B). Combining the above, we have a map α : Zp → Z(B). Since α and α correspond bijectively (follows from universal property), I only need to show there are only finitely many maps like α. Define the following action of Zp on Z(B): for a ∈ Zp and b ∈ Z(B) pick the lift (1, a) ∈ G. Define a · b = (1, a)(b, 0)(1, −a) and ϕ : Zp → Aut Z(B) as the map corresponding to this action. I will denote ab = a · b to be consistent with previous notation. Note that ab is contained in Z(B) because ϕ is the composition of ψ and the restriction map Aut B → Aut Z(B), which exists because Z(B) is characteristic in B. To avoid confusion, I will denote Z(B) multiplicatively. Let (1, a), (1, b) ∈ G. Note α(a) = α(1, a). Then we have:

α(a + b) = α(1, a + b) = σ(1, a)σ(1, b)(1, −a − b) = σ(1, a)(1, −a)(1, a)σ(1, b)(1, −b)(1, −a) = α(1, a) · (1,a)α(1, b) = α(a)aα(b).

1 1 So we see α ∈ Z (Zp, Z(B)). I need to prove that Z (Zp, Z(B)) is finite. Since Aut Z(B) is finite, we see that the kernel of ϕ must be a subgroup of finite index in Zp. Thus by lemma m 5.5 we can take m ∈ Z≥0 such that ker(ϕ) = p Zp. We have the following exact sequence:

m ϕ 0 p Zp Zp Aut Z(B).

m We see that the action of p Zp on Z(B) is trivial, hence α(a + b) = α(a)α(b) for a, b ∈ m 1 m m p Zp. This proves Z (p Zp, Z(B)) = Hom(p Zp, Z(B)), which is finite. Define r : 1 m m Z (Zp, Z(B)) → Hom(p Zp, Z(B)), f 7→ f|p Zp . We have the following exact sequence:

1 r m 0 ker(r) Z (Zp, Z(B)) Hom(p Zp, Z(B)).

m Let f ∈ ker(r). Let a ∈ Zp and x ∈ p Zp. We have: f(a + x) = f(a) · af(x) = f(a) · a1 = f(a).

m The second equality holds because f was trivial on p Zp. Thus we can define the map m 1 m f : Zp/p Zp → Z(B) which is contained in Z (Zp/p Zp, Z(B)). This proves | ker(r)| ≤

13 1 m m |Z (Zp/p Zp, Z(B))|. From lemma 5.6 we know that Zp/p Zp is finite. Since Z(B) is as 1 m well, we see that Z (Zp/p Zp, Z(B)) must be finite and ker(r) is finite as well. m Since we have an exact sequence where both ker(r) and Hom(p Zp, Z(B)) are finite, we 1 see that Z (Zp, Z(B)) is finite. This proves that there are only finitely many α, thus only finitely many α and thus ker(f1) is finite.

By lemma 5.4 and 5.7, we have a surjective map with finite kernel f2 : Aut Zp → Zp. Using lemma 5.9, we see that the image of f2 ◦ f1 has finite index in Zp. Now using lemma 5.5, this is isomorphic to Zp. Call the isomorphism f3. Now f3 ◦ f2 ◦ f1 is a surjective map. Each fi has a finite kernel, hence the composition has finite kernel as well. 

6 Automorphism groups of finite products of abelian groups

Let C be a category. Denote C0 as the collection of objects and C1 as the collection of morphisms. Let X,Y ∈ C0, I denote C(X,Y ) as the set of morphisms from X to Y and Aut X as the set of invertible morphisms from X to X.

Definition 6.1. Let C be a category. Then C is an additive category if for each X,Y ∈ C0 we have that C(X,Y ) is an abelian group, composition is bilinear and C has all finite products.

Theorem 6.2. Let C be an additive category, I a finite set and let Xi ∈ C0 for each i ∈ I. Q Q Denote Z = i∈I Xi. Then B = i,j∈I C(Xj,Xi) is a ring where addition is coordinate wise and multiplication is defined as ! X (fij)i,j∈I · (gij)i,j∈I := fik ◦ gkj , k∈I i,j∈I where we have (fij)i,j∈I , (gij)i,j∈I ∈ B with fij, gij ∈ C(Xj,Xi). Also, C(Z,Z) is a ring, with composition as multiplication, which is isomorphic as a ring to B. And Aut Z and B∗ are isomorphic as groups. The operations of B are similar to the matrix operations. I will call B the matrix ring over (Xi)i∈I , and its elements are called matrices.

Theorem 6.3. Let C be an additive category, I a finite set and let Xi ∈ C0 for each i ∈ I. Q Then Aut( i∈I Xi) is abelian iff the following statements hold: 1. ∀i ∈ I : Aut Xi is abelian, 2. ∀i, j ∈ I, i 6= j, f ∈ Aut Xi :(∀g ∈ C(Xj,Xi): fg = g) ∧ (∀g ∈ C(Xi,Xj): gf = g), 3. ∀i, j, k ∈ I, i 6= k 6= j : C(Xk,Xj)C(Xi,Xk) = 0. Q Also if Aut( i∈I Xi) is abelian, then holds: ! !   Y ∼ Y Y Aut Xi = Aut Xi Π  Hom(Xi,Xj) . i∈I i∈I i6=j∈I

I want to emphasize that I did not make a mistake in 6.3.3, I do allow i = j. Products and coproducts are equal in additive categories, see [3] page 196. This is needed in the following lemma.

14 Lemma 6.4. Let C be an additive category, I a finite set and let Xi ∈ C0 for each i ∈ I. Q Denote Z = i∈I Xi, πi : Z → X the projections and pi : Z → X the coprojections. Q 0 Let Y ∈ C0. Then ϕ : C(Y,Z) → i∈I C(Y,Xi), f 7→ (πi ◦ f)i∈I and ϕ : C(Z,Y ) → Q i∈I C(Xi,Y ), f 7→ (f ◦ pi)i∈I are isomorphisms. Proof. Note that since composition is bilinear, we see that ϕ is actually a group homomor- Q phism. If we are given (fi)i∈I ∈ i∈I C(Y,Xi), then by the categorical definition of the product we get a unique map f : Y → Z such that for each i ∈ I we have fi = πi ◦ f. It is clear by this condition that mapping such a collection to the unique map is the inverse of 0 ϕ. The proof that ϕ is an isomorphism is very similar.  Proof of theorem 6.2. Let f ∈ C(Z,Z). By applying lemma 6.4, we see that f corresponds bijectively to the family (fij : Xj → Xi)i,j∈I . Define Φ : C(Z,Z) → B, f 7→ (fij)i,j∈I . It is clear that Φ is a group isomorphism. Also Φ respects composition, which can be verified by working out the definition. Thus Φ is a ring isomorphism, from which it also follows that B is a ring. In particular, invertible elements are mapped to invertible elements. This means C(Z,Z)∗ and B∗ are isomorphic, where the isomorphism is Φ with restricted domain ∗ and codomain. Note that Aut Z is actually C(Z,Z) . 

Proof of theorem 6.3. Let B be the matrix ring over (Xi)i∈I . We know from theorem ∗ Q ∗ 6.2 that B is isomorphic to Aut( i∈I Xi). In the proof I will only use B instead of Q Aut( i∈I Xi). Note that if I is empty or contains exactly one element, the theorem is trivial. Assume |I| ≥ 2 for the rest of the proof. ⇒. I will prove the statements in order. I will constantly be defining elements of B, which I do by defining what they are on each “coordinate”(i, j). Whenever I do so, I will only specify what is different from the identity element (identity morphisms on the diagonal and zero elsewhere). This saves a lot of words. I will also be redefining a and b in every part, so they “reset”after a statement is done proving. ∗ 1. Let i ∈ I and f, g ∈ Aut Xi. Define a, b ∈ B as the elements with aii = f and bii = g. ∗ Now B is assumed to be abelian, so we have fg = (ab)ii = (ba)ii = gf. This shows that ∗ C(Xi,Xi) is abelian. ∗ 2. Let i, j ∈ I with i 6= j, f ∈ Aut Xi and g ∈ C(Xi,Xj). Redefine a, b ∈ B as elements with aii = f and bji = g. Now gf = (ba)ji = (ab)ji = g. The other statement has an analogous proof. ∗ 3. Let i, j, k ∈ I with i 6= k 6= j. Let f ∈ C(Xk,Xj) and g ∈ C(Xi,Xk). Redefine a, b ∈ B as elements with ajk = f and bki = g. Now if i 6= j we have fg = (ab)ji = (ba)ji = 0. If i = j we have id + fg = (ab)ii = (ba)ii = id, which also implies fg = 0. ∗ ⇐. Let a, b ∈ B . Write a = (aij)i,j∈I and b = (bij)i,j∈I with aij, bij ∈ C(Xj,Xi). Then we have: X (ab)ij = aik ◦ bkj, k∈I X (ba)ij = bik ◦ akj. k∈I I will need to show that ab and ba are equal on each coordinate. From 6.3.3 follows the equality (ab)ii = aiibii. From 6.3.1 we get (ab)ii = (ba)ii for each i ∈ I. Now take i, j ∈ I with i 6= j. We have: X X (ab)ij = aikbkj = bij + aij + aikbkj = bij + aij. k∈I k6=i,j

15 Note that the second equality holds because of 6.3.2 and the third equality holds because of 6.3.3. Clearly this now must be equal to (ba)ij, so a and b commute. The last statement about the isomorphism is obvious when one works out matrix multi- plication and uses conditions 1-3 to simplify the result. 

7 Abelian automorphism towers for finitely generated abelian groups

Theorem 7.1. Let A be a finitely generated abelian group. Then A has an abelian auto- morphism tower iff A is cyclic and |A| is contained in one of the following sets: m m a) {3 , 2 · 3 |m ∈ Z≥1} m m m b) {2 · 3 + 1, 2(2 · 3 + 1)|m ∈ Z≥1, 2 · 3 + 1 is prime} c) {1, 2, 4, 5, 10, 11, 22, 23, 46, 47, 94} d) {∞}

In this chapter I will refer to these four sets as Xa,Xb,Xc and Xd. Theorem 7.2. Let A be a finitely generated abelian group. Then Aut A is abelian iff A is ∼ cyclic or A = Z × Z/2Z. ∼ ∼ Proof. ⇐. We have Aut(Z × Z/2Z) = V4 and Aut Z = Z/2Z. Let n ∈ Z≥1 such that ∼ ∼ ∗ A = Z/nZ. We have End A = Z/nZ (as rings), so clearly Aut A = (End A) is abelian. ∼ r Ln ⇒. Since A is finitely generated, we have A = Z ⊕ i=1 Z/kiZ with ki | ki+1, r, n ∈ Z≥0 and if n > 0 then k1 > 1. Suppose r > 1. Then there is an injective map from Aut(Z ⊕ Z) into Aut A. We can now apply theorem 6.3 with I = 2 and X0 = X1 = Z. By statement 6.3.3 must hold Hom(Z, Z) ◦ Hom(Z, Z) = 0. Clearly this is not the case, hence Aut(Z ⊕ Z) is not abelian. This means Aut A cannot be abelian, which is a contradiction. So r ≤ 1.

Suppose n > 1. There is an injective map from Aut(Z/k1Z ⊕ Z/ak1Z) to Aut A, with a ∈ Z such that ak1 = k2. Define f : Z/ak1Z → Z/k1Z the projection map and define g : Z/k1Z → Z/ak1Z as multiplying by a. Now holds: gf(1+ak1Z) = g(1+k1Z) = a+ak1Z. Since k1 > 1 and a > 0 we have gf 6= 0. By 6.3.3, we see that Aut(Z/k1Z ⊕ Z/ak1Z) is not abelian, and neither is Aut A. This is a contradiction, so n ≤ 1. If (r, n) is (0, 0), (1, 0) or (0, 1) we see that A satisfies the theorem. Suppose (r, n) = (1, 1). ∼ This means A = Z ⊕ Z/mZ for some m ∈ Z≥2. Define π : Z → Z/mZ the canonical quotient map and f ∈ Aut(Z/mZ). Now from 6.3.2 follows fπ = π. Since π is surjective, we have that f is the identity. Thus Aut(Z/mZ) = 1, which implies m = 2. 

Definition 7.3. Define λ : Z≥1 → Z≥1 as a map of sets where λ(n) is the exponent of ∗ (Z/nZ) . This λ is called the Carmichael function. k k Proposition 7.4. Let n ∈ Z≥1. Then ϕ(n) = λ(n) iff n equals 1, 2, 4, p or 2p for an odd prime p and an integer k ∈ Z≥1. Proof. See theorem 7.3 of [4].  ∼ Lemma 7.5. Let A be a finitely generated abelian group. Then Aut A is cyclic iff A = Z ∼ or A = Z/nZ with ϕ(n) = λ(n).

16 ∼ ∗ Proof. ⇐. We have Aut Z = Z/2Z. Let n ∈ Z≥1. Note that (Z/nZ) is isomorphic to Aut Z/nZ. Also ϕ(n) = |Aut Z/nZ|. So we can conclude Aut Z/nZ is cyclic iff ϕ(n) = λ(n). ⇒. Since Aut A is cyclic, it is certainly abelian. So from theorem 7.2 follows that A =∼ ∼ Z × Z/2Z or A is cyclic. We have Aut(Z × Z/2Z) = V4, which is not cyclic. Hence A must ∼ ∼ be cyclic. Suppose A =6 Z. Take n ∈ Z≥1 such that A = Z/nZ. It is clear from the above that ϕ(n) = λ(n). 

Lemma 7.6. Let n ∈ Z≥1 such that ϕ(n) = λ(n) and ϕ(n) ∈ Xa ∪ Xb. Then n ∈ Xa ∪ Xb. Proof. From lemma 7.4 follows that n equals 1, 2, 4, pk or 2pk for an odd prime p and integer k ∈ Z≥1. The smallest integer in Xa ∪ Xb is 3 so n cannot be 1, 2 or 4. Let p be an odd k k k−1 prime and k ∈ Z≥1 such that n = p or n = 2p . In either case we have ϕ(n) = (p−1)p . m Assume ϕ(n) = 2·3 with m ≥ 1. If k > 1, we see that p must equal 3 so n ∈ Xa. Suppose k = 1. There can only be one possible prime such that ϕ sends it to 2 · 3m and that is m 2 · 3 + 1. So n ∈ Xb. Assume ϕ(n) = 2q, where q = 2 · 3m + 1 is prime and m ≥ 1. Suppose k > 1. Then we have (p − 1)pk−1 = 2q. Since there is at least one factor p, we must have p = q. However, also p − 1 = 2 must hold. This is a contradiction. For k = 1 we have p − 1 = 2q. We have: 2q + 1 = 4 · 3m + 3, which is divisible by 3. So p is not prime, which is a contradiction.

Other cases do not exist, since either ϕ(n) = 1 or 2 | ϕ(n).  ∼ 2 ∼ Proof of theorem 7.1. ⇐. We have Aut Z = Z/2Z and Aut Z = 0, so Z does indeed have an abelian automorphism tower. ∼ m Suppose A = Z/nZ for n = 2 · 3 . We have ϕ(n) = λ(n) by proposition 7.4, hence by lemma 7.5 we see that Aut A is cyclic. Now ϕ(2 · 3m) = 2 · 3m−1. We now see that Aut2 A is also cyclic. Since we keep getting powers of 3 multiplied by 2, we keep getting cyclic groups. After m+1 steps, we get the trivial group. Hence A has an abelian automorphism m tower. Note that for any odd integer n ∈ Z≥1 holds ϕ(2n) = ϕ(n). Therefore Z/3 Z also has an abelian automorphism tower. ∼ m Suppose A = Z/nZ for n = 2 · 3 + 1 prime. This prime is odd, so it satisfies proposition 7.4. Now ϕ(n) = n − 1 = 2 · 3m. We have already checked above that a cyclic group of this order has an abelian automorphism tower, hence A does as well. Since n is odd, we see that ϕ(2n) = ϕ(n). So Z/2nZ also has an abelian automorphism tower.

From proposition 7.4 and lemma 7.5 follows that for all n ∈ Xc we have Aut Z/nZ is cyclic. One can also calculate that for all n ∈ Xc we have |Aut Z/nZ| ∈ Xc. Thus Z/nZ has an abelian automorphism tower for n ∈ Xc. 2 ∼ ⇒. Note Aut (Z × Z/2Z) = Sym 3. From theorem 7.2, we can conclude that every group ∼ in the automorphism tower of A is cyclic. Assume A =6 Z. So A is a finite group and all groups in the automorphism tower of A must be finite as well. Take n ∈ Z≥1 such that ∼ A = Z/nZ. We can repeatedly use lemma 7.5 and we get the following statement: for all i ∼ i i 0 i ≥ 0 holds Aut A = Z/ϕ (n)Z, where ϕ (n) is the i-th iteration of ϕ. Note ϕ (n) = n. We also have that ϕi(n) = λi(n) for all i ≥ 0, again from lemma 7.5. i i Let n ∈ Z≥1 such that for all i ≥ 0 holds ϕ (n) = λ (n). To prove this theorem, it is enough to show that n must be contained in either Xa,Xb or Xc. Assume n 6∈ Xa ∪ Xb. The cases n = 1 and n = 2 are trivial, so also assume n ≥ 4. By repeated use of lemma i 7.6, we see that for all i ≥ 0 holds ϕ (n) 6∈ Xa ∪ Xb. I claim that there exists an i ≥ 0

17 such that ϕi(n) = 4. We know by proposition 7.4, that ϕi(n) must be 4, pk or 2pk for an odd prime p and k ≥ 1. The automorphism tower must eventually reach the trivial group, hence there is some i ≥ 0 such that ϕi(n) = 1. This implies, if we take i minimal, i−1 ϕ (n) = 2. The only way to get 2 is through 3, 6 or 4. Since 3, 6 ∈ Xa ∪ Xb, we must reach it through 4. The only thing we want to do is calculate the following set:

k k i {n ∈ Z≥4|n ∈ {4, p , 2p |p odd prime, k ∈ Z≥1}, ∃i ∈ Z≥0 : ϕ (n) = 4}.

There is no ϕ−1, but we can calculate what integers map to n under ϕ by hand for small numbers. If we work our way backwards from 4, one would not expect that we get a finite set. However, since every element must be (2 times) a prime power, we do get a finite set. The set is {5, 10, 11, 22, 23, 47, 94}. So n ∈ Xc, which completes the proof. 

8 Basic subgroups

In this chapter I will develop the theory of p-basic subgroups, which some readers may already know. I approach the theory with a definition that is less common in the litera- ture. I will also show useful properties about p-basic subgroups that are needed for the proofs in the next chapter. First I will introduce a useful convention when talking about vector spaces, after which I will need a long introduction for the definition of a p-basic subgroup.

Let K be a field, V be a K-vector space, I a set and let a : I → V, i 7→ ai be a function. (I) P Now a induces a map fa : K → V, (λi)i∈I 7→ i∈I λiai. We say a generates V when fa is surjective. We say a is linearly independent over K when fa is injective. We say that a is a basis for V if fa is an isomorphism.

Let A be an abelian group and p a prime. For n ∈ Z≥1 define A[n] := {a ∈ A : na = 0}. i Denote V = A/pA as an Fp-vector space. For i ∈ Z≥0 define Vi := im(A[p ] → A/pA, x 7→ S x + pA) and V∞ := Vi as subspaces of V . So we get the following sequence: i∈Z≥0

0 = V0 ⊆ V1 ⊆ ... ⊆ V∞ ⊆ V = A/pA.

Choose S∞ ⊂ A such that S∞ → V/V∞, s 7→ (s + pA) + V∞ is a basis for V/V∞. For i each i ∈ Z≥1 choose Si ⊂ A[p ] such that Si → Vi/Vi−1, s 7→ (s + pA) + Vi−1 is a basis for Vi/Vi−1. Define the following group:   M i (S ) (S ) B :=  (Z/p Z) i  ⊕ Z ∞ . i∈Z≥1

Note that we have |S∞| = dimFp (V/V∞) and |Si| = dimFp (Vi/Vi−1). Note that B only depends on the size of Si, so B does not depend on the choice of Si (up to isomorphism). Now define ϕ : B → A in the same manner we induced fa from a, which does depend on the choice of Si. Definition 8.1. Let A be an abelian group and p a prime. Then (B, (S ) , ϕ) is i i∈{∞}∪Z≥1 called a p-basic subgroup of A. As done more often with definitions, I will just say that B is a p-basic subgroup of A and not state Si and ϕ explicitly. Throughout this and the next chapter I will always denote the sets Si and the map ϕ whenever I talk about a p-basic subgroup. The definition becomes

18 more useful when ϕ is actually injective, so we can see B as a subgroup of A. This is actually the case. The following two theorems show some important properties of a p-basic subgroup. Theorem 8.2. Let A be an abelian group, p a prime and B a p-basic subgroup. Then ϕ is injective.

Theorem 8.3. Let A be an abelian group, p a prime, B a p-basic subgroup and n ∈ Z≥1. Denote the following subgroup of B:

M i (Si) Bn := (Z/p Z) . i∈{1,...,n}

Then the following short exact sequence is split:

ϕ|Bn 0 → Bn −−−→ A → A/ϕ(Bn) → 0.

The following theorem is a useful result and is crucial in my proofs of the two theorems above.

Theorem 8.4. Let A be an abelian group, p a prime, B a p-basic subgroup and n ∈ Z≥1. n n Then ϕ induces a map ψn : B/p B → A/p A, which is an isomorphism. Before proving these three theorems, I will prove two lemmas. These lemmas are helpful in general, since they give an idea of what the Vi look like. ∼ ∼ Lemma 8.5. We have V/V∞ = A/(pA + Ators) and for i ∈ Z>0 we have Vi/Vi−1 = i i+1 i−1 A[p ]/(pA[p ] + A[p ]), where the isomorphisms are Fp-vector space isomorphisms. Proof. First note that any isomorphism between abelian groups is also a Z-module isomor- phism. In all stated quotients we see that multiplying any element by p is always the zero element, so any group isomorphism will be an Fp-vector space isomorphism.

Define ϕ : A → V/V∞, a 7→ (a + pA) + V∞. This ϕ is a surjective map. So if I can show that the kernel is exactly pA + Ators, the first isomorphism theorem proves the first part. Let a ∈ A, then we have:

ϕ(a) = 0 ⇔ a + pA ∈ V∞ ⇔ ∃i ∈ Z≥0 : a + pA ∈ Vi i ⇔ ∃i ∈ Z≥0 : ∃b ∈ A[p ]: a + pA = b + pA i ⇔ ∃i ∈ Z≥0 : ∃b ∈ A[p ], c ∈ A : a = b + pc ⇒ ∃b ∈ Ators, c ∈ A : a = b + pc.

It is enough to show that the converse of the last implication holds. Let a ∈ pA + Ators i and write a = pc + b. Let n be the order of b. I need to find an i ∈ Z≥1, y ∈ A[p ] and c0 ∈ A such that pc + b = pc0 + y. It is enough to find a c0 ∈ A such that p(c − c0) + b has i order dividing p . To have fewer letters: it is enough to find i ∈ Z≥1 and d ∈ A such that pi(b + pd) = 0. t 0 0 Write n = p · n with p - n . Sinds Z/pZ is a field, we can take k ∈ Z≥1 such that 0 kn0−1 kn ≡ 1 mod p. Take i = t and d = p b and it works. i I will now prove the other isomorphism. Define ψ : A[p ] → Vi/Vi−1, a 7→ a + pA + Vi−1. Again ψ is surjective, so by the first isomorphism theorem it is enough to show ψ has

19 pA[pi+1] + A[pi−1] as kernel. Let a ∈ A[pi], then:

ψ(a) = 0 ⇔ a + pA ∈ Vi−1 ⇔ ∃b ∈ A[pi−1]: a + pA = b + pA ⇔ ∃b ∈ A[pi−1], c ∈ A : a = b + pc ⇐ ∃b ∈ A[pi−1], c ∈ A[pi+1]: a = b + pc. I will show that the other implication holds as well. Let a ∈ A[pi] such that ψ(a) = 0 and write a = b + pc with b ∈ A[pi−1] and c ∈ A. Now we have: 0 = pia = pi(b + pc) = pib + pi+1c = pi+1c.

i+1 Hence we have c ∈ A[p ] and we are done.  `n Lemma 8.6. For n ∈ Z≥1 we have that i=1 Si is a basis for Vn as Fp-vector space. Also ` ` Si is a basis for V∞ and S∞ q Si is a basis for V . i∈Z≥1 i∈Z≥1

Proof. I will show the first part using induction on n. Since V0 = 0, the case n = 1 is clear. Let n > 1. Now we have the following short exact sequence:

0 → Vn → Vn+1 → Vn+1/Vn → 0. Every short exact sequence of vector spaces splits, which tells us we have the following ∼ isomorphism: Vn+1 = Vn ⊕ Vn+1/Vn. From this isomorphism and induction it is evident `n that Sn+1 q i=1 Si is a basis for Vn+1.

Since V∞ is the union of all Vi, the disjoint union of the bases of Vi will give a basis for V∞. The last part follows with an analogous proof from a short exact sequence.  Proof of theorem 8.4. Consider the following diagram:

ϕ B A

πn ψ B/pnB n A/pnA

The map ψn comes from the universal property of the quotient group. So ψn is unique and n n n exists if p B is contained in the kernel of πn ◦ ϕ. Note that we have ϕ(p B) = p ϕ(B) ⊂ n p A, hence ψn exists.

I will prove this theorem by induction on n. For any n ∈ Z≥1, we have the following isomorphism:    

n ∼ M i (Si) M n (Si) n (S∞) B/p B =  (Z/p Z)  ⊕  (Z/p Z)  ⊕ (Z/p Z) . i∈{1,...,n−1} i∈Z≥n

For the base case, n = 1, we see that B/pB is a large direct sum consisting of only Fp. ` Define a : S∞ q Si → V as in lemma 8.6 such that a is a basis for V as p-vector i∈Z≥1 F space. Now ψ1 is exactly fa, so ψ1 is an isomorphism. Suppose n > 1. We have the following diagram:

ι π 0 pB/pnB 1 B/pnB 1 B/pB 0

ψn|pB ψn ψ1 ι π 0 pA/pnA 2 A/pnA 2 A/pA 0

20 The short five lemma states that if the diagram commutes, the two rows are short exact sequences and the maps ψn|pB and ψ1 are isomorphisms, then so is ψn. By the induction hypothesis we know that ψ1 is an isomorphism. It is fairly straightforward to show that both squares commute, hence I will leave this to the reader. 0 To show that ψn|pB is an isomorphism, we need to do some work. First, define A = A/A[p]. 0 0 0 0 0 Define V = A /pA and Vi and V∞ in the same manner as before with A replaced by A . 0 0 0 I claim we can choose Si = Si+1 and S∞ = S∞ to define B as a p-basic subgroup. The following proves this, using lemma 8.5:

0 0 ∼ 0 i 0 i+1 0 i−1 Vi /Vi−1 = A [p ]/(pA [p ] + A [p ]) =∼ (A[pi+1]/A[p])/((pA[pi+2]/A[p]) + (A[pi]/A[p])) =∼ A[pi+1]/(pA[pi+2] + A[pi]) ∼ = Vi+1/Vi.

0 Now if we express B in terms of Si, we have the following:   0 M i (S ) (S ) B =  (Z/p Z) i+1  ⊕ Z ∞ . i∈Z≥1

Consider the map f : A → pA, a 7→ pa. Clearly the kernel is A[p], so f induces an isomorphism between A0 and pA. Now we have:

pA/pnA = pA/pn−1(pA) =∼ A0/pn−1A0.

This isomorphism is induced by f, so again it is multiplication by p. Denote this isomor- phism by f˜ : A0/pn−1A0 → pA/pnA. In exactly the same manner, we get the isomorphism 0 n−1 0 n 0 g˜ : B /p B → pB/p B. Since A satisfies the induction hypothesis, we have ψn−1 is an isomorphism. We have the following diagram:

f˜ B0/pn−1B0 pB/pnB

ψn−1 ψn|pB g˜ A0/pn−1A0 pA/pnA

Showing that the square commutes is showing that multiplying with p before ψ is the same as multiplying after ψ. However, ψ is Z-linear so this is trivial. This shows that ψn|pB is an isomorphism since the three other maps are isomorphisms. 

Proof of theorem 8.2. Let n ∈ Z≥1, denote ψn the isomorphism as in theorem 8.4 and n πn : B → B/p B the quotient map of B. The following diagram commutes:

ϕ B A

πn ψ B/pnB n A/pnA

From the commuting property and that ψn is an isomorphism follows ker(ϕ) ⊂ ker(πn) = pnB. We even get: ker(ϕ) ⊂ T pnB. We know B explicitly and it is clear that n∈Z≥1 T pnB = 0. Hence ker(ϕ) = 0. n∈Z≥1 

21 Proof of theorem 8.3. First note that by theorem 8.2, the map ϕ is injective and we do have an exact sequence. By the splitting lemma it is enough to show there is a homomorphism r : A → Bn such that rϕ|Bn = idBn . As stated in the proof of theorem 8.4, we know that n 0 B/p B is isomorphic to a large direct sum. Now Bn is part of this direct sum, so define B 0 n as the rest of the direct sum such that fn : Bn ⊕B → B/p B is an isomorphism. Consider the following diagram:

e1 ϕ Bn B A

s e2 πB πA

0 fn n ψn n Bn ⊕ B B/p B A/p A

Take e1 and e2 the canonical embeddings and s the projection map. By definition of fn we have fn ◦ e2 = πB ◦ e1 and by definition of ψn follows that the right square commutes. −1 −1 Note that we have s ◦ e2 = idBn . Taking r = s ◦ fn ◦ ψn ◦ πA satisfies rϕ|Bn = idBn . 

9 Abelian automorphism groups

Let A be an abelian group. Then −1A will be the map −1A : A → A, a 7→ −a. Theorem 9.1. Let A be an abelian group such that Auti A is abelian for i = 0, ..., 4 and ∼ −1 Aut A is infinite. Then Aut A = Z/2Z ⊕ C where C is a Z[2 ]-module. The following theorem is trivial, but is crucial in this chapter. Definition 9.2. Let A and B be abelian groups. We say B is a direct summand of A if there exists an abelian group C such that A =∼ B ⊕ C. I use the notation B | A. Theorem 9.3. Let A and B be abelian groups with B | A and Aut A abelian. Then Aut B is abelian and Aut B | Aut A. Moreover if A has an abelian automorphism tower, then B has an abelian automorphism tower as well. Proof. The proof is obvious from theorem 6.3, which shows that we have Aut(B ⊕ C) =∼ Aut(B) ⊕ Aut(C) ⊕ Hom(B,C) ⊕ Hom(C,B).  Lemma 9.4. Let G be a group with exponent 2. Then there exists a set X such that ∼ (X) G = (Z/2Z) . Proof. Note aba−1b−1 = abab = 1, therefore G is abelian. It is enough to show that G is a vector space over the field F2, from which follows that G has a basis and we can choose X as the set consisting of all the basis elements. Choose the addition of G as a vector space the same as the operation from G. For g ∈ G define 0 · g = 0 and 1 · g = g. I leave it to the reader to verify that with these operations G is indeed a vector space over F2. 

Lemma 9.5. Let A be an abelian group such that Aut A is abelian and −1A = idA. Then A is either trivial or Z/2Z.

Proof. From −1A = idA follows that every element in A has order at most 2. From ∼ (X) lemma 9.4 we can take X a set such that A = (Z/2Z) . Suppose |X| ≥ 2. Now ∼ Aut(Z/2Z ⊕ Z/2Z) = Sym 3 must be abelian by theorem 9.3, which is a contradiction. Thus X cannot have more than one element.  −1 For a prime p denote Γp = Z[p ]/Z. For an abelian group A and n ∈ Z≥1 denote ∞ n A[n] = {a ∈ A|na = 0}. For a prime p denote A[p ] = {a ∈ A|∃n ∈ Z≥0 : p a = 0}.

22 In the following lemma and the previous notation some arbitrary prime p is used. In this chapter I will always use p = 2. Lemma 9.6. Let A be an abelian group and p a prime such that there exists a ∈ A\{0} n with pa = 0. Then Γp | A or there exists an n ∈ Z≥1 such that Z/p Z | A. Proof. We have either A[p∞] ⊂ pA or A[p∞] 6⊂ pA. I will consider both cases. ∞ 2 Suppose A[p ] ⊂ pA. Let a1 ∈ A[p]\{0}. Now a1 ∈ pA, so there exists a2 ∈ A[p ] such that pa2 = a1. We can iterate this to find ai for each i ≥ 1 such that pai = ai−1. Define −i f :Γp → A where f(p ) = ai. This map is well-defined and also injective. Consider the exact sequence, where C is the quotient:

0 → Γp → A → C → 0.

Note that Γp is a divisible group. Divisible abelian groups are injective Z-modules (see chapter 10 of [5]), meaning that they split each short exact sequence of abelian groups. Thus Γp is a direct summand of A. ∞ n Suppose A[p ] 6⊂ pA. Suppose for all n ∈ Z≥1 we have Z/p Z- A. Let B be a p-basic subgroup of A. From theorem 8.3 follows V = V∞. This can only happen if each element with order pn for some n ≥ 1 is divisible by p, or equivalently A[p∞] ⊂ pA. This is a n contradiction. Hence there exists n ∈ Z≥1 such that Z/p Z | A.  Lemma 9.7. Let A and B be non-trivial abelian groups such that Aut(A ⊕ B) is abelian. ∼ (X) Then there exists a set X such that Aut(A⊕B) = Aut(A)⊕Aut(B)⊕(Z/2Z) . Moreover, if Aut2(A ⊕ B) is abelian, then |X| ≤ 1. If also Aut3(A ⊕ B) is abelian, then |X| = 0. Proof. From theorem 6.3 we have Aut(A ⊕ B) =∼ Aut(A) ⊕ Aut(B) ⊕ Hom(A, B) ⊕ Hom(B,A). Note −1A ∈ Aut A. So for any f ∈ Hom(A, B), we have that from 6.3.2 follows −f = f. This shows 2f = 0, so the exponent of Hom(A, B) and Hom(B,A) divides ∼ (X) 2. From lemma 9.4 we can take a set X such that Hom(A, B) ⊕ Hom(B,A) = (Z/2Z) . Suppose also Aut2 (A ⊕ B) is abelian. If |X| ≥ 2 then theorem 9.3 implies that Sym 3 =∼ Aut(Z/2Z ⊕ Z/2Z) is abelian, which is a contradiction. Therefore |X| ≤ 1. Suppose also Aut3(A ⊕ B) is abelian. Let X be a set such that Aut(A ⊕ B) =∼ Aut(A) ⊕ (X) Aut(B) ⊕ (Z/2Z) . By the previous statement we know |X| ≤ 1. Assume |X| = 1. Assume that both A and B contain at least 3 elements. We see by lemma 9.5 that −1 6= id in both Aut A and Aut B. Therefore Hom(Z/2Z, Aut A) and Hom(Z/2Z, Aut B) are both (2) 2 non-zero. Also these groups have exponent 2, so we have (Z/2Z) | Aut (A ⊕ B). Now theorem 9.3 implies that Sym 3 is abelian, which is a contradiction. ∼ Suppose without loss of generality |A| = 2. If −1B = idB, then B = Z/2Z by lemma 9.5. ∼ This would mean that Aut(A⊕B) = Sym 3, which is a contradiction. Therefore −1B 6= idB n and we can use lemma 9.6 for Aut B. Assume Z/2 Z | Aut B for some n ∈ Z≥1. We can take a group C such that we have:

∼ ∼ n Aut(A ⊕ B) = Aut(Z/2Z ⊕ B) = C ⊕ Z/2 Z ⊕ Z/2Z.

n Now Aut(Z/2 Z ⊕ Z/2Z) is non-abelian by theorem 7.2. This is a contradiction, so Γ2 | Aut B. Thus we can take a group C0 such that we have:

∼ ∼ 0 Aut(A ⊕ B) = Aut(Z/2Z ⊕ B) = C ⊕ Γ2 ⊕ Z/2Z.

23 ∗ I leave it to the reader to verify that the automorphism group of Γ2 is Z2, which is isomor- phic to Z2 ⊕Z/2Z by lemma 5.7. Note that Hom(Z/2Z, Γ2) is non-trivial and has exponent (2) 2 2. Now (Z/2Z) | Aut (A ⊕ B) and from theorem 9.3 follows a contradiction. All cases lead to a contradiction, hence the assumption |X| = 1 is false. Thus |X| = 0. 

Proof of theorem 9.1. From lemma 9.5 we see −1A 6= idA, therefore Aut A has an element of order 2. This means we can use lemma 9.6 for Aut A. Suppose Γ2 | Aut A. We have (2) 2 (Z/2Z) | Aut Γ2 which is a contradiction by theorem 9.3. Let n ∈ Z≥1 such that n n ∼ n−2 Z/2 Z | Aut A. Note Aut Z/2 Z = Z/2Z ⊕ Z/2 Z for n ≥ 3. Since this is not a cyclic group, we see by theorem 7.2 that its automorphism group is not abelian. Theorem 9.3 implies a contradiction. Hence n ≤ 2. If n = 1 we have Z/2Z | Aut A, so assume n = 2. ∼ 2 Write Aut A = Z/4Z ⊕ C. Note Z/2Z | Aut A. Since Aut A is infinite, we see that C must be infinite. From lemma 9.5 follows −1C 6= idC . If there is some m ∈ Z≥1 such that m m 2 Z/2 Z | Aut C, then Aut(Z/2Z ⊕ Z/2 Z) | Aut A which is a contradiction by theorem 2 7.2 and 9.3. From lemma 9.6 follows Γ2 | Aut C. We have Γ2 ⊕ Z/2Z | Aut A. Note (2) Z/2Z | Aut Γ2 and Z/2Z | Hom(Z/2Z, Γ2), thus (Z/2Z) | Aut(Γ2 ⊕ Z/2Z). This is a contradiction by theorem 9.3. So Z/2Z | Aut A. ∼ Write Aut A = Z/2Z ⊕ C. Note that any abelian group is also a Z-module. We have that −1 C is a Z[2 ]-module iff the map f : C → C, x 7→ 2x is an isomorphism. Now f is exactly an isomorphism iff C = 2C and C[2] = 0. From theorem 6.3 we get:

2 ∼ ∼ Aut A = Aut(Z/2Z ⊕ C) = Aut(C) ⊕ Hom(C, Z/2Z) ⊕ Hom(Z/2Z,C).

However we can apply theorem 9.7 to Z/2Z ⊕ C and we see that both Hom-groups must be zero. We have Hom(C, Z/2Z) = Hom(C/2C, Z/2Z) = 0, so C = 2C. We also have Hom(Z/2Z,C) = 0, therefore C[2] = 0. 

10 References

[1] H. Wielandt, Eine Verallgemeinerung der invarianten Untergruppen, Mathematische Zeitschrift 45, 1939, 209-244 [2] R. Apon, Groups that split short exact sequences, Bachelor thesis, 2014, available at the website of Mathematical Institute of Leiden University [3] S. Mac Lane, Categories for the Working Mathematician, 2nd edition, Springer, 1978 [4] J.J. Rotman, An Introduction to the Theory of Groups, 4th edition, Springer, 1995 [5] S. Lang, Algebra, 3rd edition, Springer, 2002

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