PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 126, Number 9, September 1998, Pages 2525–2533 S 0002-9939(98)04496-7

THE MAXIMAL NORMAL p–SUBGROUP OF THE AUTOMORPHISM GROUP OF AN ABELIAN p–GROUP

JUTTA HAUSEN AND PHILLIP SCHULTZ

(Communicated by Ronald M. Solomon)

Abstract. Let p be a prime number and let G be an abelian p–group. Let Δ be the maximal normal p–subgroup of Aut G and ζ the maximal p–subgroup of its centre. Let t be the torsion radical of E(G). Then Δ = (1 + t)ζ. The result is new for p = 2 and 3, and the proof is new and valid for all primes p.

§1. Introduction

Throughout, p is an arbitrary but fixed prime and G is an abelian p–group. We let E = E(G) denote the endomorphism ring of G, Aut G its automorphism group, and Δ = O_{p}(AutG) the maximal normal p–subgroup of Aut G. The importance of Δ in determining the structure of Aut G, characterising the action of Aut G on G and recovering invariants of G from Aut G has long been recognized; see for example [3, Section 114], [6] and [1].

For p ≥ 5 , Δ considered as a group of units of E has been determined in a series of papers starting with Shoda [17] for finite groups, Freedman [2] for countable reduced groups, Hill [9] for totally projective groups and Hausen [5] for arbitrary p–groups: if t denotes the torsion radical of E (i.e. t is the ideal consisting of all elements of finite order in the Jacobson radical of E), then Δ = 1 + t .

Crucial for the proof is the following result, first stated in [5]. It was inspired by Freedman [2], stated by Leptin for k = 1 [11] and, in fact, is easily proven by induction on k following Leptin’s arguments [11, p.101]. The Pierce radical of E is defined to be

P = {ε ∈ E : p^{n}G[p]ε ≤ p^{n+1}G for all integers n ≥ 0 } .

Clearly, P is a two–sided ideal in E containing the Jacobson radical of E [13, p. 289]. Let 1 + P denote the coset containing 1 in the quotient ring E/P.

The Freedman–Leptin Lemma. Let p ≥ 5 , let G be a reduced abelian p– group, and let α = 1 + η ∈ AutG ∩ (1 + P) . Then, for any natural number k, α^{p^{k}} = 1 if and only if p^{k}η = 0 .

For p ≥ 5 , the description of Δ proceeds as follows: the ideal t is contained in the Jacobson radical of E ; since t consists of all elements in P of finite order, it follows that 1 + t is a normal p–subgroup of Aut G and thus is contained in Δ [3, Theorem 114.4]. Conversely, Δ induces the identity mapping in all Ulm factors

\begin{matrix}c \\ ◯ \end{matrix}1998 American Mathematical Society

Received by the editors January 28, 1997.

1991 Mathematics Subject Classification. Primary 20K10, 20F28, 20K30.

p^{n}G[p]/p^{n+1}G[p] where n is a non-negative integer [2], so that Δ ≤ 1 + P , and the Freedman–Leptin Lemma proves Δ = 1 + t .

Two new phenomena make this approach impossible for the two smallest primes. First, the maximal p–subgroup of the center ζ Aut G of Aut G is contained in Δ . While [ζ Aut G]_{p} = 1 if p is odd, [ζ Aut G]_{2} = ⟨−1⟩ if p = 2. But if G is an unbounded 2–group, multiplication by −2 is an endomorphism in P of infinite order, so −1 is not in 1 + t.

Second, while for primes p ≥ 5 the set of all automorphisms in 1 + P of finite order forms a group, in fact a p–group which must be normal in Aut G and hence equals 1 + t , we shall show by examples that this need not be the case when p = 2 or p = 3 . For these primes, there exist p–groups G having an automorphism α of order p for which α − 1 ∈ P has infinite additive order. This answers in the negative a problem posed by Leptin [11, p. 102]. Moreover, for p = 2 , our example will be such that α ∉ (1 + t) ⟨−1⟩.

Nevertheless, we are able to prove the following theorem, which is valid for all primes:

Theorem. Let p be a prime, let G be an abelian p–group, and let Δ be the maximal normal p–subgroup of the automorphism group of G. Then Δ = (1+t)ζ, where ζ denotes the maximal p–subgroup of the centre of Aut G. Thus, Δ = 1+t if p ≥ 3 or G is bounded, and Δ = (1 + t) × ⟨−1⟩ if G is an unbounded 2–group.

It is interesting to note that if G has an unbounded basic subgroup, then as an abstract ring, t determines G up to isomorphism; furthermore, every automorphism of t is induced by an automorphism of G [8]. One may well conjecture that the same is true of Δ. If true, this would be a far–reaching extension of the results of Leptin [10] and Liebert [12].

The strategy of the proof may be outlined as follows: it is easy to show that 1 + t ≤ Δ (2.6). Of course, ⟨−1⟩ ≤ Δ when p = 2 . For the reverse inclusion, we first show that Δ ⊆ 1 + P and assume the existence of some α ∈ Δ such that α − 1 (and, for p = 2, −α − 1, too) has infinite order. We then use an elaborate construction to produce mixed commutators of the form [γ, α] (and [β, −α] if p = 2) which have infinite order. Since they would have to belong to Δ, this is a contradiction.

Any unexplained notation can be found in the standard references [3] for abelian groups and [15] for general group theory. In particular, [x, y] = x^{−1}y^{−1}xy.

§2. Preliminaries

We begin with the promised examples which show that for p = 2 and p = 3 no substitute for the Freedman–Leptin Lemma can be hoped for. The examples also illustrate a simple case of the strategy outlined above.

Examples 2.1. For each prime p ∈ {2, 3}, there exist an abelian p–group G and an automorphism α = 1 + η ∈ 1 + P such that α^{p} = 1 but η has infinite additive order. When p = 2, η+2 also has infinite order. In addition, there exists β ∈ Aut G such that [β, α] has infinite multiplicative order.

Proof. If p = 2 , let G = ⨁_{i∈ℕ}(⟨a_{i}⟩ ⊕ ⟨b_{i}⟩), where a_{i} has order 2^{i+1} and b_{i} has order 2^{i}. For each i, define a_{i}η = b_{i} and b_{i}η = −2b_{i}. Then η ∈ P , 2^{k}η ≠ 0 and 2^{k}(η + 2) ≠ 0 for all positive integers k. Furthermore, η^{2} + 2η = 0 .

MAXIMAL NORMAL p–SUBGROUP 2527

If p = 3 , let G = ⨁_{i∈N}(⟨a_{i}⟩ ⊕ ⟨b_{i}⟩ ⊕ ⟨c_{i}⟩), where a_{i}, b_{i} and c_{i} have orders 3^{3i}, 3^{3i−1} and 3^{3i−2} , respectively. For each i, define a_{i}η = b_{i}, b_{i}η = c_{i}, and c_{i}η = −3c_{i} − 3b_{i}. Then η ∈ P , 3^{k}η ≠ 0 for all k, and η^{3} + 3η^{2} + 3η = 0 .

In either case, let α = 1 + η. Then α^{p} = 1 , so α ∈ AutG ∩ (1 + P) , and α ∉ (1 + t)ζ.

We construct the automorphism β only for the case p = 2, the case p = 3 being similar. For each i ≥ 1, define

a_{i}ε = 0 and b_{i}ε = \begin{cases}a_{i−1} & if i is even, \\ 0 & if i is odd. \end{cases}

Then ε^{2} = 0, so β = 1 + ε ∈ Aut G with inverse 1 − ε. Since α^{−1} = α and εηε = 0, a routine calculation shows that [β, α] = 1 + ψ, where ψ = ηε + εη + ηεη. But ψ^{2} = 0, while ψ has infinite additive order. Hence [β, α]^{k} = 1 + kψ for all k ∈ ℕ, so [β, α] has infinite multiplicative order.

We shall need some results of elementary number theory. The first one is well known.

Lemma 2.2. Let r, n and k be positive integers such that n is not divisible by p and p^{r}n ≤ p^{k}. Then

(1) The maximal p–power dividing \binom{p^{k}}{p^{r}·n} is p^{k−r}.

(2) If

M = M(p, k, r, n) = ∑_{j=1}^{p^{k}} \binom{p^{k}}{j}(p^{r} n)^{j} ,

then p^{k+r} divides M , and if p^{k+r+1} divides M , then p = 2 and r = 1 .

Proof. The first assertion is easily proved in the spirit of [15, p. 39]. In order to verify the second, let j be an integer such that 1 ≤ j ≤ p^{k}, and let x_{j} = \binom{p^{k}}{j}p^{rj}n^{j} . By (1), x_{j} is divisible by p^{k+rj} if (j, p) = 1 . Suppose j = p^{t}m, with t and m positive integers and m relatively prime to p. By (1), p^{k−t+rj} divides x_{j} , and k − t + rp^{t}m = k + r + r(p^{t}m − 1) − t ≥ k + r since r(p^{t}m − 1) ≥ 2^{t} − 1 ≥ t. Thus, each x_{j} is divisible by p^{k+r}. Suppose that p^{k+r+1} divides M . Since p^{k+r+1} does not divide x_{1} , there must exist j ≥ 2 such that p^{k+r+1} does not divide x_{j} , and j cannot be relatively prime to p. Thus, j = p^{t}m with t and m as above, and k − t + rj = k + r. It follows that

t = r(j − 1) = r(p^{t}m − 1) ≥ r(2^{t} − 1) ≥ 2^{t} − 1 .

Since t ≥ 1 , this implies t = 1 = r and p = 2 .

The maximal normal p–subgroups of the automorphism groups of elementary and of divisible groups are known:

Lemma 2.3 ([2], [5]). Let A be an elementary abelian and D a divisible p–group. Then O_{p}(AutA) = 1 , and O_{p}(AutD) = 1 unless p = 2, in which case O_{2}(AutD) = ⟨−1⟩.

Proof. The first assertion is Lemma 4.8 of [2]. Let D be divisible. By Theorem 6.5 of [5], every normal torsion subgroup of AutD is contained in the centre, which consists of the multiplications by p–adic units [3, 115.1]. It is well known that the p–component of the group of p–adic units is trivial except when p = 2, in which case it is cyclic of order two generated by −1 [16, II.3.8].

Corollary 2.4. Let G = H ⊕ D, where H is reduced and D is divisible. Let δ ∈ Δ . Then δ − 1 ∈ P , and D(δ − 1) = 0 unless p = 2, in which case δ|_{D} ∈ ⟨−1⟩.

Proof. Let n be a nonnegative integer. Every automorphism of p^{n}G[p]/p^{n+1}G[p] is induced by an automorphism of G [3, 114.1]; clearly, every automorphism of D is induced by an automorphism of G. Thus, the result follows from Lemma 2.3.

From now on we shall assume that G = H ⊕ D, where H is reduced and D is divisible.

The following subset of P will play an important rôle. It contains t , and equals t if G is reduced. Let

Φ = { ε ∈ P | p^{m}Gε ≤ D for some m ∈ ℕ and Dε = 0 } .

Lemma 2.5. Let ε ∈ Φ and let m be a natural number such that p^{m}Gε ≤ D. Then ε^{2m+3} = 0 , and (1 + ε)^{p^{n}} = 1 + p^{n}ε for all integers n ≥ 3m + 2 .

Proof. The hypotheses imply Gε ≤ G[p^{m}] + D. Let \overline{G} = G/D and let \overline{ε} be the endomorphism of \overline{G} induced by ε. Then p^{m}\overline{ε} = 0, so that

\overline{G}\overline{ε} ≤ \overline{G}[p^{m}] ≤ \overline{G}[p^{m+1}] = (G[p^{m+1}] + D)/D .

The restriction of \overline{ε} to \overline{G}[p^{m+1}] is in P(\overline{G}[p^{m+1}]), which equals the Jacobson radical of E(\overline{G}[p^{m+1}]) since \overline{G}[p^{m+1}] is bounded [7]. By [14], JE(\overline{G}[p^{m+1}]) is nilpotent of class at most 2m + 1 . Thus, (\overline{ε} |_{\overline{G}[p^{m+1}]})^{2m+1} = 0, which implies G[p^{m+1}]ε^{2m+1} ≤ D. Hence,

Gε^{2m+3} ≤ (G[p^{m}] + D)ε^{2m+2} ≤ Dε = 0 .

Let n be an integer such that n ≥ 3m + 2 . Then

(1 + ε)^{p^{n}} = 1 + p^{n}ε + ∑_{j=2}^{p^{n}} \binom{p^{n}}{j}ε^{j} = 1 + p^{n}ε + ∑_{j=2}^{2m+2} \binom{p^{n}}{j}ε^{j} .

Fix 2 ≤ j ≤ 2m + 2 . By hypothesis, p^{m}ε^{2} = 0 . Let j = p^{r}·q for some integers r and q with p and q relatively prime. By Lemma 2.2, \binom{p^{n}}{j}ε^{j} = p^{n−r}ℓε^{j} for some integer ℓ not divisible by p. Suppose n−r < m. Then r > n−m ≥ 3m+2−m = 2m + 2, which implies

j = p^{r}q ≥ 2^{r} > r > 2m + 2 .

Thus, \binom{p^{n}}{j} ε^{j} = 0 .

While our Examples 2.1 demonstrate that the only if part of the Freedman-Leptin Lemma is lost in the general case, we do have the following weak version of the if part.

Proposition 2.6. Let ε ∈ Φ . The following conditions are equivalent:

(1) (1 + ε)^{p^{k}} = 1 for some positive integer k.

(2) ε ∈ t .

(3) 1 + ε ∈ Δ . In particular, 1 + t ≤ Δ .

Proof. Let ε ∈ Φ and let m ∈ ℕ be such that p^{m}Gε ≤ D. By 2.5, (1 + ε)^{p^{n}} = 1 + p^{n}ε for all integers n ≥ 3m + 2 . Assume (1), and choose n ≥ max(k, 3m + 2) . Then 1 = 1 + p^{n}ε, which implies (2). Conversely, assume (2). Then p^{ℓ}ε = 0 for some natural number ℓ. Choose n ≥ max(ℓ, 3m + 2) . Then (1 + ε)^{p^{n}} = 1 . Thus,

(1) and (2) are equivalent. Since t is an ideal of E contained in the Jacobson radical of E , it follows that 1 + t is a normal p–subgroup of Aut G. Hence, 1 + t ≤ Δ , and (2) implies (3). Trivially, (3) implies (1).

Corollary 2.7. For every prime p and every p–group G, (1 + t)ζ ≤ Δ .

§3. The reduced case

The following result will be crucial for the proof of our theorem. It shows that the construction of Examples 2.1 can be extended to deal with arbitrary p–groups.

Proposition 3.1. Let G be a reduced abelian p–group and let α = 1 + η ∈ Δ . Then η ∈ t , or p = 2 and 2 + η ∈ t .

The proof will be by contradiction using an elaborate construction. Throughout, we shall assume the following

Situation and Notations: The p–group G is reduced, α = 1+η is an element of Δ , and k ∈ ℕ is a fixed integer such that α^{p^{k}} = 1 . Then

∑_{j=1}^{p^{k}} \binom{p^{k}}{j}η^{j} = 0 . (3.2)

We let σ = α^{−1} − 1 . By Corollary 2.4

η , σ ∈ P (3.3)

and

η + σ + ησ = 0 , ησ = ση . (3.4)

Also, we let α′ = −α and put η′ = α′ − 1 , and σ′ = (α′)^{−1} − 1 . Then

η′ = −(2 + η) , σ′ = −(2 + σ) . (3.5)

The primed elements will come into play only when p = 2 . Since the entire construction would have to be repeated, we consider them in any case.

We also fix a basic subgroup B of G and a canonical set S of (nonzero) independent generators. Thus B = ⨁_{s∈S}⟨s⟩ = ⨁_{i∈ℕ} B_{i}, where B_{i} is generated by the elements in S of order p^{i}. For each n ∈ ℕ, put B^{(n)} = ⨁_{i=1}^{n} B_{i} and G_{n} = p^{n}G + ⨁_{i≥n+1} B_{i}. Then, for each n, G = B^{(n)} ⊕ G_{n} [13, p. 280]. Choose a family {π_{s}}_{s∈S} of canonical projections such that tπ_{s} = δ_{st}s, t ∈ S, and G_{n}π_{s} = 0 if s ∈ B^{(n)} . Define G_{0} = G.

It will be convenient to write e(x) = n if x ∈ G and o(x) = p^{n}.

Lemma 3.6. Assume the situation and notations above and, furthermore, assume that both η and η′ have infinite order. Then there exist natural numbers u′_{i}, u_{i}, v_{i} and elements a′_{i}, a_{i}, b_{i} in G belonging to the canonical set of generators of B such that, for each i ∈ ℕ, the following hold:

(1) e(a′_{i}) = u′_{i} , e(a_{i}) = u_{i} , e(b_{i}) = v_{i} , (2) e(a′_{i}η′) ≤ u′_{i} < 3u′_{i} < e(a_{i}η) ≤ u_{i} < 3u_{i} < v_{i},

(3) v_{i−1} < e(a′_{i}η′) (put v_{0} = 0 ),

(4) if π_{i} : G → ⟨b_{i}⟩ denotes the canonical projection to ⟨b_{i}⟩ and r_{i}, m_{i}, r′_{i}, m′_{i} are integers such that b_{i}σπ_{i} = p^{r_{i}}m_{i}b_{i} , b_{i}σ′π_{i} = p^{r′_{i}}m′_{i}b_{i}, and m_{i}, m′_{i} are not divisible by p, then r_{i} ∉ [u_{i} − 2u_{i−1} − k , u_{i} − 2u_{i−1} − 1] and r′_{i} ∉ [u′_{i} − 2u′_{i−1} − k , u′_{i} − 2u′_{i−1} − 1] (put u_{0} = u′_{0} = 0 ).

Proof. We claim that, for each natural number n, there exist u′_{i}, u_{i}, v_{i} and canonical generators a′_{i}, a_{i}, b_{i} of B , i = 1, …, n, satisfying (1) through (4). The proof is by induction on n. Since the proof for n = 1 is almost identical to the inductive step from n to n + 1 , it is left to the reader. Assume the claim holds for n. Since p^{v_{n}}Gη′ ≠ 0 , there exists i such that p^{v_{n}}B_{i}η′ ≠ 0 , and there is a canonical generator x of B_{i} with v_{n} < e(xη′) . Now p^{3i+k}Gη′ ≠ 0 ; thus there exist j and a canonical generator y of B_{j} such that 3i + k < e(yη′) . Proceeding this way, there exist ℓ and m and canonical generators z and w of B_{ℓ} and B_{m}, respectively, such that

\begin{aligned}v_{n} < e(xη′) ≤ e(x) = i & < 3i + k < e(yη′) ≤ e(y) = j \\ & < 3j < e(zη) < 3ℓ + k < e(wη) ≤ e(w) = m . \end{aligned}

Since p^{3m}G ≠ 0, it is possible to choose v_{n+1} > 3m such that B_{v_{n+1}} ≠ 0 . Let b_{n+1} be a canonical generator of B_{v_{n+1}} . Then e(b_{n+1}) = v_{n+1} . Let π_{n+1} denote the canonical projection from G onto ⟨b_{n+1}⟩. Let r′_{n+1} , r_{n+1} , m′_{n+1} , m_{n+1} be integers such that

b_{n+1}σ′π_{n+1} = p^{r′_{n+1}}m′_{n+1}b_{n+1} and b_{n+1}σπ_{n+1} = p^{r_{n+1}}m_{n+1}b_{n+1},

where m′_{n+1} , m_{n+1} are not divisible by p. If r′_{n+1} ∉ [i − 2u′_{n} − k , i − 2u′_{n} − 1] , let a′_{n+1} = x. By construction, if r′_{n+1} ∈ [i − 2u′_{n} − k , i − 2u′_{n} − 1] , then r′_{n+1} ∉ [j −2u′_{n} −k, j −2u′_{n} −1] . In this case, let a′_{n+1} = y. Similarly, r_{n+1} cannot belong to both the intervals [ℓ − 2u_{n} − k , ℓ − 2u_{n} − 1] and [m − 2u_{n} − k , m − 2u_{n} − 1] . Choose a_{n+1} to be either z or w depending on which is the case. One verifies that the augmented sets of integers and elements in G satisfy the requirements of the lemma.

We shall define two endomorphisms ε and ε′ as follows: for each i, define b_{i}ε_{i} = a_{i} and let ε_{i} = π_{i}ε_{i}. Since G_{v_{i}}π_{i} = 0 , each ε_{i} is an endomorphism of G. Let ε = ∑_{i∈ℕ} p^{u_{i−1}}ε_{i}. By [13, 6.3], ε ∈ E . Similarly, define b_{i}ε′_{i} = a′_{i}, ε′_{i} = π_{i}ε′_{i}, and let ε′ = ∑_{i∈ℕ} p^{u′_{i−1}}ε′_{i} ∈ E .

Lemma 3.7. Assume the hypotheses of 3.6. Then, for every integer i ≥ 0 , G_{v_{i}}[p^{u_{i+1}}]ε = 0 = G_{v_{i}}[p^{u_{i+1}}]ε′ (Reminder: v_{0} = 0 and G_{0} = G).

Proof. Let i ≥ 0 be an integer and let x ∈ G_{v_{i}}[p^{u_{i+1}}] . Then xπ_{j} = 0 for j ≤ i. Let j ≥ i + 1 . Then xπ_{j} = p^{v_{j}−u_{i+1}}n_{j}b_{j} for some integer n_{j} . Thus, xπ_{j}ε = p^{v_{j}−u_{i+1}+u_{j−1}}n_{j}a_{j} which is zero since v_{j} − u_{i+1} > 3u_{j} − u_{i+1} ≥ 2u_{j} . Thus, xε = ∑_{j∈ℕ} p^{u_{j−1}}xπ_{j}ε_{j} = 0 . A similar argument holds for ε′.

Corollary 3.8. For every ϕ ∈ E , εϕε = ε′ϕε′ = 0 . In particular, ε^{2} = 0 = (ε′)^{2} .

Proof. Since G is reduced, it suffices to show the maps annihilate B, which will be the case if they annihilate every b_{i}. Let ϕ ∈ E . Note that b_{i}εϕ = p^{u_{i−1}}a_{i}ϕ ∈ p^{u_{i−1}}B^{(v_{i−1})} ⊕ G_{v_{i−1}}[p^{u_{i}}] . The definition of ε together with Lemma 3.7 implies b_{i}εϕε = 0 . The identical argument shows that b_{i}ε′ϕε′ = 0 .

We are ready for the

Proof of Proposition 3.1. Assume the situation and notations above. Also, assume by way of contradiction that the conclusion of 3.1 is false. Then η has infinite additive order, and when p = 2, so does η′. It follows that G is unbounded. Hence, no matter what prime p was given, both η and η′ have infinite order, which means the previous auxiliary results are available.

Using Corollary 3.8, the maps γ = 1 + ε and γ′ = 1 + ε′ are automorphisms of G. Define δ = γ^{−1}α^{−1}γα and δ′ = (γ′)^{−1}(α′)^{−1}γ′α′. Then δ ∈ Δ and, if p = 2 , so is δ′. An easy calculation using (3.4) and 3.8 shows that

δ = 1 + ψ with ψ = εη + σε + σεη and ψ^{2} = 0 ,

and similarly,

δ′ = 1 + ψ′ with ψ′ = ε′η′ + σ′ε′ + σ′ε′η′ and (ψ′)^{2} = 0 .

Thus, for every n ∈ ℕ, δ^{n} = 1 + nψ. Since δ ∈ Δ , δ^{p^{ℓ}} = 1 and hence p^{ℓ}ψ = 0 for some ℓ ∈ N . Choosing ℓ sufficiently large, we also have p^{ℓ}ψ′ = 0 if p = 2 . Fix i such that both u_{i−1} > ℓ and u′_{i−1} > ℓ. Then p^{u_{i−1}}ψ = 0 and, provided p = 2 , p^{u′_{i−1}}ψ′ = 0 . Hence

0 = p^{u_{i−1}}b_{i}ψ = p^{u_{i−1}}b_{i}εη + p^{u_{i−1}}b_{i}σε + p^{u_{i−1}}b_{i}σεη . (3.9)

Since b_{i}σ ∈ B^{(v_{i})} ⊕ G_{v_{i}}[p^{v_{i}}] , Lemma 3.7 implies that b_{i}σ = ∑_{j=1}^{i} k_{j}b_{j} + x, where the k_{j} are integers and xε = 0 . From (3.3) we conclude that p divides k_{i} and, by 3.6(4),

k_{i} = p^{r_{i}}m_{i} with r_{i} ≥ 1 , (p, m_{i}) = 1 . (3.10)

It follows that p^{u_{i−1}}b_{i}σε = p^{2u_{i−1}}k_{i}a_{i} and p^{u_{i−1}}b_{i}σεη = p^{2u_{i−1}}k_{i}a_{i}η. From (3.9) and b_{i}εη = p^{u_{i−1}}a_{i}η we deduce that (1 + k_{i})p^{2u_{i−1}}a_{i}η = −k_{i}p^{2u_{i−1}}a_{i}. Let n be an integer such that n(1 + k_{i}) ≡ −m_{i}(mod p^{u_{i}}) . By (3.10), p does not divide n and (p^{2u_{i−1}}a_{i})η = p^{r_{i}}n · p^{2u_{i−1}}a_{i}, which by construction is nonzero. Hence, r_{i} < u_{i} −2u_{i−1} . Also, for every positive integer j, (p^{2u_{i−1}}a_{i})η^{j} = (p^{r_{i}}n)^{j} ·p^{2u_{i−1}}a_{i}. It follows from (3.2) that

p^{2u_{i−1}} [∑_{j=1}^{p^{k}} \binom{p^{k}}{j}(p^{r_{i}}n)^{j}] a_{i} = p^{2u_{i−1}}M(p, k, r_{i}, n)a_{i} = 0 . (3.11)

Suppose either that p ≠ 2 , or that p = 2 and r_{i} ≥ 2 . Then, by Lemma 2.2, the highest p–power dividing M(p, k, r_{i}, n) is p^{k+r_{i}} . Thus, 3.11 implies that k + r_{i} ≥ u_{i} − 2u_{i−1} ; above we saw that r_{i} < u_{i} − 2u_{i−1} . It follows that

r_{i} ∈ [u_{i} − 2u_{i−1} − k , u_{i} − 2u_{i−1} − 1],

contrary to our delicate construction 3.6(4). Consequently, we must have that p = 2 and r_{i} = 1 . In this situation, the primed elements and maps have the identical properties. We have (δ′)^{2^{ℓ}} = 1 , (1 + ψ′)^{2^{ℓ}} = 1 + 2^{ℓ}ψ′ = 1 , and the choice of i was such that

0 = 2^{u′_{i−1}}b_{i}ψ′ = 2^{u′_{i−1}}b_{i}ε′η′ + 2^{u′_{i−1}}b_{i}σ′ε′ + 2^{u′_{i−1}}b_{i}σ′ε′η′ .

Let the k′_{j} be integers such that b_{i}σ′ = ∑_{j=1}^{i} k′_{j}b_{j} + x′ with x′ε′ = 0 . As before, letting n′ be an integer such that n′(1 + k′_{i}) ≡ −m′_{i}(mod 2^{u′_{i}}) , we have that n′ is odd and, for every positive integer j, 2^{2u′_{i−1}}a′_{i}(η′)^{j} = (2^{r′_{i}}n′)^{j}2^{2u′_{i−1}}a′_{i}. By construction, r′_{i} < u′_{i}−2u′_{i−1} . As before, it follows that 2^{2u′_{i−1}}M(2, k, r′_{i}, n′)a′_{i} = 0 . Observing (3.5) and the fact that r_{i} = 1 , we conclude that

2^{r′_{i}}m′_{i}b_{i} = b_{i}σ′π_{i} = −b_{i}(2 + σ)π_{i} = −(2 + 2^{r_{i}}m_{i})b_{i} = −2(1 + m_{i})b_{i} .

But m_{i} is odd, hence r′_{i} ≥ 2 . It follows from Lemma 2.2 that k + r′_{i} ≥ u′_{i} − 2u′_{i−1} . Thus, r′_{i} belongs to the interval [u′_{i} −2u′_{i−1} −k , u′_{i} −2u′_{i−1} −1] , contrary to 3.6(4). This final contradiction concludes the proof.

Proposition 3.1 yields the most difficult part of our theorem:

Corollary 3.12. If G is a reduced p–group, then Δ ≤ (1 + t)ζ.

§4. Proof of the Theorem

As before, we let G = H ⊕ D, where H is reduced and D is divisible. By Corollary 2.7, it suffices to show Δ ≤ (1 + t)ζ. By 3.12, this is the case when G is reduced. Thus, we may assume that D ≠ 0 . Let δ = 1 + η ∈ Δ . Since Δ induces a normal p–subgroup of Aut(G/D) , it follows from 3.12 that there exists a natural number m such that either p^{m}Gη ≤ D, or p = 2 and 2^{m}G(2 + η) ≤ D. If the latter is the case, replace δ by −δ = 1−(2+η) . Thus, we may assume without loss of generality that p^{m}Gη ≤ D. By Corollary 2.4, η ∈ P . If Dη = 0 , then η ∈ Φ , and δ ∈ 1 + t , by Proposition 2.6. Suppose Dη ≠ 0 . It follows from Corollary 2.4 that p = 2 and δ|_{D} = −1 , the multiplication by −1 . Choose any θ ∈ Hom(H, D) and extend it to an endomorphism of G by defining Dθ = 0 . Then θ^{2} = 0 and γ = 1 + 2^{m}θ ∈ Aut G. Let α = δγδ^{−1}γ^{−1} . Then α ∈ Δ and α|_{D} = 1 . As before, let δ^{−1} = 1 + σ. Then η + σ + ησ = 0 . Let x ∈ G. Since Gη and Gσ both are contained in G[2^{m}] + D, on which γ acts as the identity, we have

xα = xδγδ^{−1}γ^{−1} = x + 2^{m}xθσ + xσ + xη + xησ,

which implies x(α − 1) = −2^{m+1}xθ. It follows that α − 1 ∈ Φ , and Proposition 2.6 implies that θ ∈ t . We have shown that Hom(H, D) is a torsion group, which, since D ≠ 0 , implies 2^{n}H = 0 for some positive integer n. But then 2 + η ∈ Φ and −δ = 1 − (2 + η) ∈ 1 + t , completing the proof.

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Department of Mathematics, University of Houston, Houston, Texas 77204-3476 E-mail address: hausen@uh.edu

Department of Mathematics, University of Western Australia, Nedlands 6009, Australia

E-mail address: schultz@maths.uwa.edu.au