2.3 The Krull-Schmidt Theorem

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1 2.3. THE KRULL-SCHMIDT THEOREM The Krull-Schmidt Theorem Every finitely generated abelian group is a direct sum of finitely many indecomposable abelian groups Z and Z p n. We will study a large class of (not necessarily abelian) groups which can be decomposed into a direct product of indecomposable groups. Def. A group G is indecomposable if G {e} and G is not the direct product of two of its proper subgroups. Every simple group is indecomposable (e.g. A n for n 4, Z p for prime p). However indecomposable groups need not be simple (e.g. Z, Z p n for n 2, and S n ) Def. Let G be a group. G satisfies the ascending chain condition (ACC) if for every chain G 1 < G 2 < of normal subgroups of G there is an integer n such that G i = G n for all i > n; G satisfies the descending chain condition (DCC) if for every chain G 1 > G 2 > of normal subgroups of G there is an integer n such that G i = G n for all i > n. Ex. Every finite group satisfies both chain conditions. Ex. (HW) Z satisfies ACC but not DCC. Ex. (HW) Z(p ) = {a/b Q/Z b = p n for some n N} satisfies DCC but not ACC. Thm If G satisfies either ACC or DCC, then G is a direct product of a finite number of indecomposable subgroups. Proof. Call a group good if it is a direct product of a finite number of indecomposable subgroups; call it bad otherwise. Facts: 1. Indecomposable groups are good. 2. If P = H K and both H and K are good, then P is good. Suppose on the contrary, G is bad. Then G is not indecomposable. So G = H 1 K 1 for some H 1, K 1 G. At least one of H 1 and K 1 should be bad. Without lost of generality, assume that K 1 is bad. Then K 1 = H 2 K 2 for H 2, K 2 K 1, where at least one of H 2 and K 2 is bad, say K 2 is bad. Repeating the process, we have G = H 1 (H 2 (H 3 ) ) H 1 H 2 H 3 There are an infinite strictly ascending chain and an infinite strictly descending chain of G: H 1 < H 1 H 2 < H 1 H 2 H 3 < and G > K 1 > K 2 > These violate both ACC and DCC. Therefore G is good.

2 42 CHAPTER 2. THE STRUCTURE OF GROUPS Def. An endomorphism f of a group G is called a normal endomorphism if af(b)a 1 = f(aba 1 ) for all a, b G. Lem If ϕ and ψ are normal endomorphisms of a group G, then so is ϕ ψ. 2. If ϕ is a normal endomorphism of G and if H G, then ϕ(h) G. 3. If ϕ is a normal automorphism of a group G, then ϕ 1 is also normal. Def. If ϕ and ψ are endomorphisms of G, then ϕ + ψ : G G is the function defined by x ϕ(x)ψ(x). For general group G, ϕ+ψ is not necessarily commutative on +, and it is not necessarily a homomorphism. Lem Let G = H 1 H m have projections π i : G H i and inclusion ι i : H i G. Then the sum of any k distinct ι i π i is a normal endomorphism of G. Moreover, the sum of all the ι i π i is the identity function on G. Proof. If ϕ = k i=1 ι iπ i, then ϕ(h 1,, h m ) = h 1 h k, that is, ϕ = ιπ, where π is the projection of G onto the direct factor H 1 H k and λ is the inclusion of H 1 H k into G, It follows that ϕ is a normal endomorphism of G and, if k = m, that ϕ = 1 G. Lem Let G have both chain conditions. If ϕ is a normal endomorphism of G, then ϕ is an injection if and only if it is a surjection. (Thus, either property ensures that ϕ is an automorphism.) Proof. Suppose that ϕ is an injection and that g ϕ(g). Then ϕ n is an injection, and thus ϕ n (g) ϕ n (ϕ(g)) = ϕ n+1 (G). There is a stricly descending chain of subgroups G > ϕ(g) > ϕ 2 (G) >. Now ϕ normal implies ϕ n is normal. By Lemma 2.14, ϕ n (G) G for all n, and so the DCC is violated. Therefore, ϕ is a surjection. Assume that ϕ is a surjection. Then ϕ n is a surjection for all n. We claim that Ker ϕ = {e}. Otherwise, there is x e and x Ker ϕ. Write x = ϕ n (g) for some g G. Then g Ker ϕ n but g Ker ϕ n+1. Thus we have a strictly ascending chain of normal subgroups {e} = Ker ϕ 0 < Ker ϕ 1 < Ker ϕ 2 <. So the ACC is violated. Thus ϕ is an injection.

3 2.3. THE KRULL-SCHMIDT THEOREM 43 Lem 2.17 (Fitting s Lemma). Let G have both chain conditions and ϕ a normal endomorphism of G. Then G = Ker ϕ n Im ϕ n for some n 1. Proof. Since ϕ is normal, we have the following two chains of normal subgorups G Im ϕ 1 Im ϕ 2 and {e} Ker ϕ 1 Ker ϕ 2 By hypothesis there is an n such that Im ϕ k = Im ϕ n and Ker ϕ k = Ker ϕ n for all k n. Claim: Ker ϕ n Im ϕ n = {e}. Otherwise, there is x e and x Ker ϕ n Im ϕ n. Let x = ϕ n (g). Then ϕ 2n (g) = ϕ n (x) = e. So g Ker ϕ 2n = Ker ϕ n. So x = ϕ n (g) = e, a contradiction. Thus Ker ϕ n Im ϕ n = {e}. Given x G, ϕ n (x) = ϕ 2n (g) by Im (ϕ n ) = Im (ϕ 2n ). So x = kϕ n (g) for some k Ker ϕ n. This shows that G = Ker ϕ n Im ϕ n. Therefore G = Ker ϕ n Im ϕ n. Def. An endomorphism ϕ of a group G is nilpotent if ϕ n (G) = {e} for some n. Cor If G is an indecomposable group having both chain conditions, then every normal endomorphism ϕ of G is either nilpotent or an automorphism. Proof. There is n 1 such that G = Ker ϕ n Im ϕ n. Then either Ker ϕ n = {e} or Im ϕ n = {e} since G is indecomposable. If Ker ϕ n = {e}, then Ker ϕ = {e}, and thus ϕ is an injection, and thus ϕ is an automorphism by Lemma If Im ϕ n = {e}, then ϕ is nilpotent. Lem Let G be an indecomposable group with both chain conditions, and let ϕ and ψ be normal nilpotent endomorphisms of G. If ϕ + ψ is an endomorphism of G, then it is normal nilpotent. Proof. a(ϕ + ψ)(b)a 1 = aϕ(b)a 1 + bψ(b)a 1 = ϕ(aba 1 ) + ψ(aba 1 ) = (ϕ + ψ)(aba 1 ). If ϕ + ψ is an endomorphism, then it is normal. By Lemma 2.18, ϕ + ψ is either nilpotent or an automorphism. If ϕ + ψ is an automorphism, then Lemma 2.14 says that its inverse γ is also normal. One has 1 G = (ϕ + ψ)γ = ϕγ + ψγ where ϕγ := ϕ γ. Let λ = ϕγ and µ = ψγ. Then 1 G = λ + µ. So x 1 = λ(x 1 )µ(x 1 ) and, taking inverses, x = µ(x)λ(x); that is, λ + µ = µ + λ. The equation λ(λ + µ) = (λ + µ)λ implies that λµ = µλ. Then for any integer m > 0, (λ + µ) n = ( ) m λ i µ m i. (2.2) i i Nilpotence of ϕ and ψ implies nilpotence of λ = ϕγ and µ = ψγ (they cannot be automorphisms because they have nontrivial kernels); there are r, s Z + with λ r = 0 and µ 2 = 0. If m = r + s 1, then (2.2) shows that 1 G = (λ + µ) m = 0, forcing G = {e}. This is a contradiction, for every indecomposable group is nontrivial.

4 44 CHAPTER 2. THE STRUCTURE OF GROUPS Cor Let G be an indecomposable group having both chain conditions. If ϕ 1,, ϕ n is a set of normal nilpotent endomorphisms of G such that every sum of distinct ϕ s is an endomorphism, then ϕ ϕ n is nilpotent. Proof. Induction on n. Thm 2.21 (Krull-Schmidt). Let G be a group having both ACC and DCC. If G = H 1 H s = K 1 K t are two decompositions of G into indecomposable factors, then s = t and there is a reindexing of K s so that H i K i for all i. Moreover, given any r between 1 and s, the reindexing may be chosen so that G = H 1 H r K r+1 K s. Remark. The last conclusion is stronger than saying that the factors are determined up to isomorphism; one can replace factors of one decomposition by suitable factors from the other. Proof. We shall give the proof when r = 1; the reader may complete the proof for general r by inducton. Given the first decomposition, we must find a reindexing of the K s so that H i K i for all i and G = H 1 K 2 K t. Let π i : G H i and ι i : H i G be the projection and inclusion from G = H 1 H s, and π i : G K i and ι i : K i G the projection and inclusion from G = K 1 K t. The maps ι i π i and ι i π i are normal endomorphisms of G. By Lemma 2.15, every partial sum ι i π i is a normal endomorphism of G. Hence, every partial sum of 1 H1 = π 1 ι 1 = π 1 1 G ι 1 = π 1 ( ι jπ j)ι 1 = π 1 ι jπ jι 1 is a normal endomorphism of H 1. Since 1 H1 is not nilpotent and H 1 satisfies both chain conditions (as G does and every normal subgroup of H 1 is also normal in G), Corollary 2.20 gives an index j with π 1 ι j π j ι 1 an automorphism. We reindex so that π 1 ι 1 π 1 ι 1 is an automorphism of H 1 ; let γ be its inverse. We claim that π 1 ι 1 : H 1 K 1 is an isomorphism. We have (γπ 1 ι 1 )(π 1 ι 1) = 1 H1. Let us show that θ = (π 1 ι 1)(γπ 1 ι 1 ) = 1 K 1. First, θ 2 = θ. Second, 1 H1 = 1 H1 1 H1 = (γπ 1 ι 1)(π 1ι 1 )(γπ 1 ι 1)(π 1ι 1 ) = (γπ 1 ι 1)θ(π 1ι 1 ), so that θ 0. Were θ nilpotent, then θ 2 = θ forces θ = 0. Therefore, θ is an automorphism of K 1, and so θ 2 = θ gives θ = 1 K1. Thus π 1 ι 1 : H 1 K 1 is an isomorphism. Now π 1 sends K 2 K t into {e} while π 1 ι 1 restricts to an isomorphism on H 1. Therefore H 1 (K 2 K t ) = {e}. Then G := H 1, K 2 K t = H 1 K 2 K t.

5 2.3. THE KRULL-SCHMIDT THEOREM 45 If x G, then x = k 1 k 2 k t, where k j K j. Since π 1 ι 1 is an isomorphism, the map β : G G, defined by x π 1 ι 1 (k 1)k 2 k t, is an injection with image G. By Lemma 2.16, β is a surjection; that is, G = G = H 1 K 2 K t. Finally, K 2 K t G/H 1 H 2 H s, so that the remaining uniqueness assertions follow by induction on max{s, t}.

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