# Automorphisms of Regular Wreath Productp-Groups

Authors: Jeffrey M. Riedl
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

We present a useful new characterization of the automorphisms of the regular wreath product groupP of a finite cyclic p-group by a finite cyclic p-group, for any prime p, and we discuss an application. We also present a short new proof, based on representation
theory, for determining the order of the automorphism group Aut(P), where P is the regular wreath product of a finite cyclic p-group by an arbitrary finite p-group.


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## Body

## 1. Introduction

LetP denote the regular wreath product group C≀Q, where Q is an arbitrary nontrivial finite p-group, for some prime p, and where C is an any finite cyclic p-group. Thus P is the semidirect product B⋊Q, where B is a direct product of |Q| copies of C, and where Q acts via automorphisms on B by regularly permuting these direct factors.In [1], Houghton determines some information on the structure of the automorphism group Aut(P). Using this work of Houghton (see also [2, Chapter 5]), it is possible to calculate the order of Aut(P). Our first result in this paper is to present an alternative method for calculating the order of Aut(P). Our approach to this calculation is to apply the Automorphism Counting Formula (established in [3]), a general formula for the order of the automorphism group Aut(G) of a monolithic finite group G in terms of information about the complex characters of G and information about how G is embedded as a subgroup of a particular finite general linear group. A finite group is said to be monolithic if and only if  it has a unique minimal normal subgroup. Thus a finite p-group is monolithic if and only if  its center is cyclic. Let |C|=pe and |Q|=pn. Throughout this paper we assume that pen≥3, which excludes only the case where p=2 and e=n=1, for which P is dihedral of order 8.Theorem 1.1.
Aut(P) has order |Aut(Q)|(p-1)pa, where a=2epn-e-1.Because the dihedral group of order8 has an automorphism group of order 8, the condition pen≥3 is a necessary hypothesis for Theorem 1.1.The next result is a step along the way to proving Theorem1.1. We mention it here.Theorem 1.2.
Letq be any prime-power larger than 1 such that pe is the full p-part of q-1. Then the general linear group GL(pn,q) has exactly one conjugacy class of subgroups whose members are isomorphic to P.Now suppose that the groupQ of order pn is cyclic. Since Aut(Q) has order (p-1)pn-1, Theorem 1.1 yields |Aut(P)|=(p-1)2p2epn+n-e-2. Using knowledge of |Aut(P)| and little more than an elementary counting argument, we obtain a useful new characterization of the automorphisms of P. Before stating this characterization, we establish some notation.Hypothesis 1.3.
Assume that the groupQ of order pn is cyclic. Let x0,x1,…,xpn-1 be a collection of elements of order pe that constitutes a generating set for the homocyclic group B of exponent pe and of rank pn. Let w be a generator for the cyclic group Q and suppose that xuw=xu-1 for each u∈{1,…,pn-1} and that x0w=xpn-1.Under Hypothesis1.3, it is clear that {xpn-1,w} is a generating set for the group P, and so every automorphism of P is determined by where it maps these two elements.Neumann [4] has characterized the regular wreath product groups (including infinite groups) for which the so-called base group is a characteristic subgroup. This general result of Neumann implies that B is always a characteristic subgroup of P for the particular class of wreath product groups P considered in this paper. Nevertheless, in our proof of Theorem 1.1 we present our own brief argument (see Step 7) that B is a characteristic subgroup of P. From this fact it follows that [B,P] is a characteristic subgroup of P.We are now ready to state the main result of this paper.Theorem 1 A.
Assume Hypothesis1.3. Then the group B/[B,P] is cyclic of order pe, and therefore has a unique maximal subgroup which one denotes as D/[B,P], and so D is a characteristic subgroup of P that satisfies |B:D|=p. Let ℰ denote the set of all elements g∈P of order pn that satisfy the condition P=〈B,g〉. Then for each pair of elements (a,b) such that a∈B-D and b∈ℰ, there exists an automorphism of P that maps xpn-1 to a and maps w to b. Furthermore, every automorphism of P is of this type.In the notation of Theorem A, the information that we have about the subgroupD and the set ℰ makes it clear that every automorphism of P maps the set B-D to itself and maps the set ℰ to itself. It is not difficult to see that the element xpn-1 belongs to the set B-D and that the element w belongs to the set ℰ. From this perspective, we might summarize Theorem A as stating that every mapping that could possibly be an automorphism of P actually is an automorphism of P.Theorem A gives us a factorization ofA=Aut(P), namely, A=CA(w)CA(x′) with CA(w)∩CA(x′)=1, where x′=xpn-1. Houghton's main result in [1] is a factorization of A, namely, A=CA(w)I⋊Q* with CA(w)∩I=1, where I denotes the group of inner automorphisms of P induced by elements of B, and where Q* is the image of the usual embedding of Aut(Q) in A (see [2]). In particular Q*≅Aut(Q). Since I⊆CA(x′), these two factorizations are the same if and only if  Q*⊆CA(x′). However, Q* permutes the elements x0,x1,…,xpn-1 with x′=xpn-1 lying in a regular orbit, and so Q*∩CA(x′)=1. Hence these two factorizations are the same if and only if  Q*=1, which happens only when |Q|=2.We now discuss an application of Theorem A. In [5] we classify up to isomorphism the nonabelian subgroups of the wreath product group P=ℤpe≀ℤp for an arbitrary prime p and positive integer e such that pe≥3. In [6] we use the characterization of the elements of A=Aut(P) that is provided by Theorem A to compute the index |NA(H):CA(H)| for each group H of class 3 or larger appearing in this classification. For each such group H, we then observe that this index is equal to the order of the automorphism group Aut(H), from which we deduce that the group NA(H)/CA(H) is isomorphic to Aut(H), which says that the full automorphism group Aut(H) is realized inside the group A=Aut(P).In Section3 we prove Theorems 1.1 and 1.2. In Section 4 we prove Theorem A. In Section 2 we discuss some preliminary results used in our proof of Theorem 1.1.LetIrr(G) denote the set of irreducible ordinary characters of a finite group G.


## 2. Preliminaries

For each finite groupG and prime-power q, let mindeg(G,q) denote the smallest positive integer m such that the general linear group GL(m,q) contains a subgroup that is isomorphic to G. Thus mindeg(G,q) is the minimal degree among all the faithful F-representations of the group G, where F denotes the field with q elements. For any groups H and G such that H⊆G, we have mindeg(H,q)≤mindeg(G,q).Definition 2.1.
LetG be a monolithic finite group, let q be a prime-power that is relatively prime to the order of G, and let m=mindeg(G,q). We say that the ordered triple (G,q,m) is a monolithic triple in case every faithful irreducible ordinary character of G has degree at least m. Assuming that (G,q,m) is a monolithic triple, we define ℱ(G,q) to be the set of all faithful irreducible ordinary characters of G of degree m. We say that the monolithic triple (G,q,m) is good provided that every value of each character belonging to the set ℱ(G,q) is a ℤ-linear combination of complex (q-1)st roots of unity.The following is a special case of result that was proved in [3]. We call this result the Automorphism Counting Formula. It is the key to establishing Theorem 1.1.Theorem 2.2.
Let(G,q,m) be a good monolithic triple. Suppose that Γ=GL(m,q) has a unique conjugacy class of subgroups whose members are isomorphic to G. Let H be any subgroup of Γ that is isomorphic to G. Then |Aut(G)|(q-1)=|ℱ(G,q)|·|NΓ(H)|.In our proof of Theorem1.1, the idea is to define a good monolithic triple (G,q,m) with G=P that satisfies the hypothesis of Theorem 2.2. The conclusion of Theorem 2.2 would then yield |Aut(G)| provided that we know in advance |ℱ(G,q)| and |NΓ(H)|.Given a monolithic groupG, in order to define a good monolithic triple (G,q,m) we must choose an appropriate prime-power q and then calculate mindeg(G,q). The following result may be used to calculate mindeg(G,q) for certain groups G and prime-powers q.Lemma 2.3.
LetG be any finite group containing an abelian p-subgroup B of exponent pe and of rank r, where p is a prime. Let F be any field containing a primitive peth root of unity. If there exists a faithful F-representation of G of degree r, then mindeg(G,F)=r.Proof.
The hypotheses yieldmindeg(B,F)≤mindeg(G,F)≤r. It remains to show that r≤mindeg(B,F). The hypothesis on F implies that every irreducible F-representation of B has degree 1 and that the characteristic of the field F is not p. Let 𝒳 be any faithful F-representation of B, and let n be its degree. By Maschke's theorem, 𝒳 is similar to a faithful F-representation 𝒴 consisting of diagonal matrices. Let E be the subgroup of GL(n,F) consisting of all diagonal matrices of order dividing pe. Then 𝒴(B)⊆E while E is homocyclic of exponent pe and of rank n. Since 𝒴 is faithful, indeed 𝒴(B) is an abelian p-group of rank r. It follows that r≤n. Therefore mindeg(B,F)≥r, as desired.One of the hypotheses of Theorem2.2 is that the general linear group GL(m,q) has a unique conjugacy class of subgroups whose members are isomorphic to G. The following result (Lemma 4.5 in [3]) is useful for establishing this condition in certain situations.Lemma 2.4.
LetF be a field containing a primitive peth root of unity, where p is some prime and e is some positive integer. Let G be any finite group containing an abelian normal p-subgroup B of exponent pe and of rank r. Then every faithful F-representation of G of degree r is similar to a representation 𝒴 such that 𝒴(B) consists of diagonal matrices and 𝒴(G) consists of monomial matrices.Using Theorem2.2 to calculate the order of the automorphism group Aut(G) for a given monolithic triple (G,q,m) requires that we know in advance the cardinality of the set ℱ(G,q) that was defined in Definition 2.1. The following result is helpful for calculating the cardinality of the set ℱ(G,q) in certain situations.Lemma 2.5.
Letp be a prime and let P be a monolithic finite p-group. One defines the set 𝒜={ψ∈Irr(P)∣ψisfaithful}. Let n be a nonnegative integer and suppose that every character belonging to the set 𝒜 has degree pn. Then |𝒜|=|P|(p-1)/p2n+1.Proof.
We define the setℬ=Irr(P)-𝒜. Let N be the unique minimal normal subgroup of P, and note that ℬ={ψ∈Irr(P)∣N⊆kerψ}. Hence the set ℬ may be identified with the set Irr(P/N). We have |N|=p, and so |P/N|=|P|/p. By Corollary 2.7 in [7], along with the fact that Irr(P)=𝒜∪ℬ is a disjoint union, we deduce that
(2.1)|P|=∑ψ∈𝒜ψ(1)2+∑ψ∈ℬψ(1)2=|𝒜|p2n+|P|p.
Solving this equation for |𝒜|, we obtain the desired conclusion.Using Theorem2.2 to calculate the order of the automorphism group Aut(G) for a given monolithic triple (G,q,m) requires that we know in advance the order of the normalizer of a certain subgroup H in the general linear group GL(m,q). The following result (which is part of Theorem 4.4 in [3]) is useful for this task in certain situations.Theorem 2.6.
LetΓ=GL(m,q) where q>1 is any prime-power and m is any positive integer. Let F be the field with q elements, let F0 be any nontrivial subgroup of the multiplicative group F×=F-{0}, and let E be the group of all diagonal matrices in Γ having the property that each entry along the diagonal belongs to F0. Let S be the subgroup of Γ consisting of all permutation matrices, and note that S≅Sym(m). Let T be any transitive subgroup of the symmetric group S and let H=E⋊T. If E is a characteristic subgroup of H, then |NΓ(H)|=|NS(T):T|·|H|(q-1)/|F0|.The following rather specialized result will be used in our proof of Theorem1.1.Lemma 2.7.
Letp be any prime and let e, n, and j be positive integers such that j≤n. Then the condition epn-j(pj-1)≤j holds if and only if  p=2 and e=n=j=1.Proof.
First, an easy inductive argument shows that2j-1>j whenever j≥2. Now suppose that epn-j(pj-1)≤j holds. First we show that j=1. Assuming instead that j≥2, we get pj-1≥2j-1>j, forcing epn-j(pj-1)>j, a contradiction. Hence j=1, and so epn-1(p1-1)≤1, which forces each of the positive integers e, pn-1, and p-1 to be 1. Therefore e=n=1 and p=2, as desired. The reverse implication is trivial.The next two results on permutation groups will be used later in this article.Lemma 2.8.
LetH1 and H2 be isomorphic transitive subgroups of order n of the symmetric group Sym(n). Then H1 and H2 are conjugate subgroups of Sym(n).Proof.
For eachα∈Ω={1,…,n} and each x∈Sym(n), let α·x denote the image of α under x. For i∈{1,2}, the maps fi:Hi→Ω defined by fi(x)=1·x are bijections. Let θ:H1→H2 be an isomorphism. The composition y=f2θf1-1:Ω→Ω is an element of Sym(n). It suffices to show that y-1xy=θ(x) for each x∈H1. A straightforward calculation (left to the reader) yields α·y-1xy=α·θ(x) for arbitrary α∈Ω.Theorem 2.9.
LetH be any transitive subgroup of order n in the symmetric group S=Sym(n). Then the normalizier NS(H) is isomorphic to the holomorph H⋊Aut(H).The following basic lemma is needed for our proof of Theorem2.9.Lemma 2.10.
LetG be a group of permutations of a set Ω, let H be a transitive subgroup of G, and let C=CG(H). For each α∈Ω, the stabilizer subgroup Cα is trivial.Proof.
Letx∈Cα. To prove that x=1, it suffices to show that β·x=β for arbitrary β∈Ω, since G acts faithfully. There exists h∈H such that α·h=β. Since x∈C, we have hx=xh, and so β·x=(α·h)·x=α·(hx)=α·(xh)=(α·x)·h=α·h=β.Proof of Theorem2.9.
LetG be a group that is isomorphic to H. Let V=G⋊A where A=Aut(G). First we identify a subgroup D of V that is isomorphic to G and that centralizes G. The rule x↦φxx-1 defines an injective homomorphism θ:G→V, where φx∈A is the inner automorphism induced by x. Let D=θ(G). For x, y∈G, observe that
(2.2)θ(x)-1yθ(x)=(xφx-1)y(φxx-1)=x(φx-1yφx)x-1=x(x-1yx)x-1=y.
Next we embedV as a subgroup of S in such a way that G becomes a transitive (in fact regular) subgroup of S. Since coreV(A)=1, the action of V on the set Ω consisting of the right cosets of A in V is faithful. We now argue that the action of G on Ω is regular. Since |G|=|Ω|, it suffices to show that each nonidentity element of G fixes no element of Ω. Let x∈G and Av∈Ω such that x fixes Av. Thus Avx=Av and so vxv-1∈A. Since x∈G⊲V, we obtain vxv-1∈A∩G=1, and so x=1, as desired. Now label the members of Ω as the numbers 1,2,…,n. In this way we regard V as a subgroup of S.
SinceH and G are isomorphic transitive subgroups of order n in S, by Lemma 2.8 we may complete the proof by showing that NS(G)=V. Write C=CS(G). Lemma 2.10 implies that every orbit in the action of C on {1,…,n} has size |C|. Hence |C| divides n=|G|=|D|. But since D centralizes G, we have D⊆C. It follows that D=C.
WriteN=NS(G). By the N-Mod-C Theorem, the integer |N|/|C| divides |A|, which says that |N| divides |C|·|A|. Recalling that |C|=|D|=|G|, this says that |N| divides |G|·|A|=|V|. But since G⊲V⊆S, we have V⊆N. It follows that V=N.


## 3. Proof of Theorem1.1

Let{xu∣u∈Q} be a collection of elements of order pe that constitutes a generating set for the homocyclic group B of exponent pe and of rank |Q|=pn. We now define an action of the group Q on the set {xu∣u∈Q}. For each pair u, v∈Q, we let xuv=xuv, where the product uv is computed in Q. This action naturally gives rise to an action of Q via automorphisms on the group B. Let P=B⋊Q denote the semidirect product group corresponding to this action. Let ℱ denote the set consisting of all functions from Q into the additive group ℤpe. For each function f∈ℱ, we define the element
(3.1)x(f)=∏u∈Qxuf(u)∈B.
Each element of B has the form x(f) for some unique function f∈ℱ. We define the element z∈B of order pe by letting z denote the product of all the elements xu for u∈Q.Step 1.
For each subgroupL of Q, the centralizer CB(L) is equal to the set of all elements x(f) such that the function f∈ℱ is constant on each of the left cosets of L in Q.Proof.
LetT be a transversal for the left cosets of L in Q. For each t∈T, observe that the set {xu∣u∈tL} is an orbit in the action of L on the set of generators {xu∣u∈Q} for B.Step 2.
The groupP is monolithic, and its center is the cyclic group 〈z〉 of order pe.Proof.
SinceB is abelian and the action of Q via automorphisms on B is faithful, the center of P=B⋊Q is CB(Q). By Step 1, CB(Q) is the cyclic group generated by the element z. Finally, since P is a p-group whose center is cyclic, P is indeed monolithic.Following standard notation (see [7]), we define the inertia subgroup of any character θ∈Irr(B) as the subgroup IP(θ)={x∈P∣θx=θ}.Step 3.
For each characterθ∈Irr(B) such that IP(θ)>B, every irreducible constituent of the induced character θP is not faithful.Proof.
For each pair of functionsf, g∈ℱ we define the dot product f·g to be the value
(3.2)f·g=∑u∈Qf(u)g(u)∈ℤpe.
Let ϵ be any primitive complex peth root of unity. For each function g∈ℱ, we define the character φg∈Irr(B) by φg(x(f))=ϵf·g for every function f∈ℱ. It is clear that every irreducible ordinary character of B is of the form φg for some function g∈ℱ.
Letθ∈Irr(B) such that IP(θ)>B. Since kerθP is equal to the intersection of the kernels of the irreducible constituents of θP, it suffices to show that kerθP>1. Because P=B⋊Q, we have IP(θ)=B⋊L for some nontrivial subgroup L of Q. Let T be any transversal for the left cosets of L in Q. Since 1<L⊆Q, the prime p divides |L|. Since θ∈Irr(B), we have θ=φg for some function g∈ℱ. Because the character θ is L-invariant, the function g must be constant on each left coset of L in Q. This says that for each t∈T, there exists a value ct∈ℤpe such that g(u)=ct for each element u∈tL.
By Step2, 〈zpe-1〉 is the unique minimal normal subgroup of P. Note that zpe-1=x(f) for the constant function f∈ℱ defined as f(u)=pe-1 for u∈Q. Observe that
(3.3)θ(zpe-1)=ϵf·gwheref·g=∑u∈Qf(u)g(u)=∑t∈T∑u∈tLf(u)g(u).
For each t∈T, using the fact that |tL|=|L| is divisible by p, we deduce that
(3.4)∑u∈tLf(u)g(u)=∑u∈tLpe-1ct=|L|pe-1ct=0.
It follows that f·g=0, which yields zpe-1=x(f)∈kerθ. Hence 〈zpe-1〉⊆kerθ. Using kerθP=coreP(kerθ) and 1<〈zpe-1〉⊲P, we obtain 1<〈zpe-1〉⊆kerθP, as desired.We define the set𝒜={ψ∈Irr(P)∣ψisfaithful}.Step 4.
For each characterχ∈𝒜 we have χ(1)=pn, and for each element x∈P the value χ(x) is a sum of complex peth roots of unity. Furthermore |𝒜|=(p-1)|P|/p2n+1.Proof.
Letχ∈𝒜 be arbitrary and let θ∈Irr(B) be any irreducible constituent of the restriction χB. Hence χ is an irreducible constituent of the induced character θP. Since B⊆IP(θ) and since χ is faithful, Step 3 yields IP(θ)=B. By the Clifford Correspondence [7, Theorem  6.11], it follows that θP is irreducible, and so χ=θP. Since θ∈Irr(B) while B is abelian, we have θ(1)=1. Therefore χ(1)=θP(1)=|P:B|θ(1)=|Q|=pn.
Sinceχ=θP with θ∈Irr(B) and B⊲P, the character χ vanishes off B. Furthermore, because B is an abelian p-group of exponent pe, every value of θ is a complex peth root of unity. By Theorem 6.2 in [7], the restriction χB is a sum of conjugates of θ in P. Hence for each element x∈B, the value χ(x) is a sum of complex peth roots of unity.
Finally, Lemma2.5 yields |𝒜|=|P|(p-1)/p2n+1, as desired.Letq>1 be any prime-power such that pe is the full p-part of q-1. Let Γ=GL(pn,F) where F is the field with q elements. Let D, S, and M denote the subgroups of Γ consisting of all diagonal matrices, permutation matrices, and monomial matrices, respectively. Note that M=D⋊S and that S is isomorphic to the symmetric group of degree pn. Let E denote the subgroup of Γ consisting of all diagonal matrices of order dividing pe. Thus E is homocyclic of exponent pe and of rank pn. Note that E is the unique Sylow p-subgroup of the abelian group D, and that E is a separator subgroup of Γ.We will now define a faithful representation𝒵:P→Γ. Recall that {xu∣u∈Q} is a collection of elements of order pe that constitutes a generating set for the homocyclic group B of exponent pe and of rank |Q|=pn. We index the rows and the columns of the matrices in Γ by the elements of the group Q. We choose an arbitrary element ω of order pe in the cyclic multiplicative group of nonzero elements in the field F. For each u∈Q, we define 𝒵(xu) to be the diagonal matrix in Γ whose (u,u)-entry is ω, and each of whose other diagonal entries is 1. Thus 𝒵(B)=E consists of diagonal matrices. We define 𝒵|Q:Q→Γ to be the right regular representation of the group Q. Thus 𝒵(Q) consists of permutation matrices and is a regular subgroup of the symmetric group S. The action of Q by conjugation on B inside the group P is similar to the action of 𝒵(Q) by conjugation on 𝒵(B) inside the group Γ. Thus, since P=QB and B∩Q=1, we have a faithful representation 𝒵:P→Γ whose image 𝒵(P)=𝒵(Q)𝒵(B) is a subgroup of SE.Step 5.
mindeg(P,F)=pn.Proof.
Recall that𝒵 is a faithful F-representation of P of degree pn; use Lemma 2.3.The next step establishes Theorem1.2.Step 6.
Every faithfulF-representation of P of degree pn is similar to 𝒵.Proof.
By Lemma2.4, every faithful F-representation of P of degree pn is similar to a faithful F-representation 𝒳 such that 𝒳(B)⊆D and 𝒳(P)⊆M. Since E is the unique Sylow p-subgroup of D, indeed 𝒳(B)⊆E. Since 𝒳 is faithful, the p-groups 𝒳(B) and E are homocyclic of exponent pe and of rank pn. It follows that 𝒳(B)=E. That E is the unique Sylow p-subgroup of D yields E⊲NΓ(D). Satz II.7.2(a) in [8] yields NΓ(D)=M, so E⊲M=DS. Let R be a Sylow p-subgroup of S. Thus ER is a Sylow p-subgroup of M. Since 𝒳(P) is a p-subgroup of M, Sylow's theorem asserts that 𝒳 is similar (by a matrix in M) to a representation 𝒴 such that 𝒴(P)⊆ER. We have 𝒴(B)=E, since E⊲M. Thus 𝒴(P)/E and 𝒵(P)/E are regular subgroups of the symmetric group ES/E≅Sym(pn), and are both isomorphic to Q. By Lemma 2.8, conjugation by some element of ES/E maps 𝒴(P)/E to 𝒵(P)/E. Conjugation by the unique preimage of this element under the natural isomorphism S→ES/E maps 𝒴(P) to 𝒵(P). Hence 𝒴 is similar to 𝒵.Step 7.
B is a characteristic subgroup of P.Proof.
We argue thatB is the only abelian normal subgroup of index pn in P. Let A be an abelian normal subgroup of P such that |P:A|=pn and A≠B. Write |AB:B|=pj with j∈{1,…,n} and let L=AB∩Q. We now argue that AB=B⋊L. Since L⊆Q while B∩Q=1, we have B∩L=1. Because B⊆AB, Dedekind's lemma yields BL=AB∩BQ=AB∩P=AB, and so AB=B⋊L. From this we obtain |L|=|AB:B|=pj. Since A and B are abelian, we have A∩B⊆Z(AB). It follows that A∩B⊆CB(L)⊆B and |B:CB(L)|≤|B:A∩B|=pj. By Step 1, we have |CB(L)|=(pe)|Q:L|=pepn-j. Since |B|=pepn, it follows that |B:CB(L)|=pepn-j(pj-1). Thus epn-j(pj-1)≤j. By Lemma 2.7, this contradicts the hypothesis pen≥3.Step 8.
The normalizerNΓ(𝒵(P)) has order (q-1)|P|·|Aut(Q)|/pe.Proof.
UsingP=B⋊Q and E=𝒵(B), we obtain 𝒵(P)=E⋊𝒵(Q). By Step 7 and the fact that 𝒵 is faithful, E=𝒵(B) is a characteristic subgroup of 𝒵(P). Since 𝒵(Q) is a regular subgroup of the symmetric group S and since 𝒵(Q)≅Q, Theorem 2.9 implies that the normalizer NS(𝒵(Q)) is isomorphic to the holomorph of Q. Therefore |NS(𝒵(Q)):𝒵(Q)|=|Aut(Q)|. The statement now follows from Theorem 2.6.Step 9.
|Aut(P)|=(p-1)|Aut(Q)|p2epn-e-1.Proof.
By Steps2, 4, and 5, (P,q,pn) is a good monolithic triple and ℱ(P,q)=𝒜. Thus Step 4 yields |ℱ(P,q)|=(p-1)|P|/p2n+1. By Step 6, 𝒵(P) belongs to the unique conjugacy class of subgroups of Γ whose members are isomorphic to P. In view of Step 8, Theorem 2.2 yields |Aut(P)|=(p-1)|Aut(Q)|·|P|2/pe+2n+1 where |P|=pepn+n.


## 4. Proof of Theorem A

Assume Hypothesis1.3. Let ℱ denote the set of all functions from the set 𝒰={0,1,…,pn-1} into the additive group ℤpe. For each function f∈ℱ, we define the element
(4.1)x(f)=x0f(0)x1f(1)⋯xpn-1f(pn-1)∈B.
Each element of B has the form x(f) for some unique f∈ℱ. The mapping φ:B→ℤpe defined by φ(x(f))=f(0)+f(1)+⋯+f(pn-1) is a surjective homomorphism. Hence B/kerφ is cyclic of order pe. To establish Theorem A, our first task is to prove that B/[B,P] is cyclic of order pe. For this it suffices to show that [B,P]=kerφ.Lemma 4.1.
For each functionf∈ℱ, the commutator element [x(f),w] has the form
(4.2)x0f(1)-f(0)x1f(2)-f(1)⋯xpn-2f(pn-1)-f(pn-2)xpn-1f(0)-f(pn-1).Proof.
Note that[x(f),w]=x(f)-1x(f)w. Conjugating x(f) by w, we obtain
(4.3)x(f)w=(x0w)f(0)(x1w)f(1)(x2w)f(2)⋯(xpn-1w)f(pn-1)=xpn-1f(0)x0f(1)x1f(2)⋯xpn-2f(pn-1)=x0f(1)x1f(2)⋯xpn-2f(pn-1)xpn-1f(0).
Since x(f)-1=x0-f(0)x1-f(1)⋯xpn-2-f(pn-2)xpn-1-f(pn-1), the result follows.Theorem 4.2.
[B,P]=kerφ.Proof.
Let[B,w] denote the subgroup of P that is generated by all elements of the form [b,w] with b∈B. Using Q=〈w〉, we can show that [B,w]=[B,Q]. Since P=BQ while B is abelian, it is clear that [B,Q]=[B,P]. Hence it suffices to show that [B,w]=kerφ.
To show that[B,w]⊆kerφ, we must verify that [x(f),w]∈kerφ for each f∈ℱ, but this is obvious by Lemma 4.1. Next we argue that kerφ⊆[B,w]. An arbitrary element of kerφ has the form x(g) for some function g∈ℱ satisfying g(0)+g(1)+⋯+g(pn-1)=0. To establish that x(g)∈[B,w], we will now define a particular function f∈ℱ such that [x(f),w]=x(g). Let f(0)=0, and for each u∈{1,…,pn-1} let f(u)=g(0)+g(1)+⋯+g(u-1). It follows that for each u∈{0,1,…,pn-2} we have f(u+1)-f(u)=g(u). Furthermore, using the condition g(0)+g(1)+⋯+g(pn-1)=0, we obtain
(4.4)f(0)-f(pn-1)=0-∑v=0pn-2g(v)=g(pn-1).
By Lemma 4.1, we deduce that [x(f),w]=x(g). Therefore x(g)∈[B,w], as desired.The cyclic groupℤpe has a unique subgroup of index p, namely, pℤpe={pa∣a∈ℤpe}. Let D be the group consisting of all those elements x(f) in B such that φ(x(f))∈pℤpe. It is clear that kerφ⊆D⊆B and |B:D|=p.Corollary 4.3.
kerφ and D are characteristic subgroups of P.Proof.
By Step7 in the proof of Theorem 1.1, B is a characteristic subgroup of P. It follows that [B,P] is a characteristic subgroup of P. By Theorem 4.2, we deduce that kerφ is a characteristic subgroup of P. Since B/kerφ is cyclic, D is the only subgroup of P that satisfies the conditions kerφ⊆D⊆B and |B:D|=p. Because B and kerφ are characteristic subgroups of P, it follows that D is a characteristic subgroup of P.For the next result, we need a formula (due to Philip Hall) for raising the product of two group elements to an arbitrary positive integer power. For any positive integern and any elements a and b belonging to some group, the element (ab)n may be written as
(4.5)an(a-(n-1)ba(n-1))(a-(n-2)ba(n-2))⋯(a-2ba2)(a-1ba1)b.
This says that (ab)n=anban-1ban-2⋯ba2ba1b. Furthermore, in case all the conjugates of b by powers of a commute with each other (which is automatically true if b is contained in an abelian normal subgroup of any group containing a and b), this formula becomes
(4.6)(ab)n=anbba1ba2⋯ban-2ban-1=an∏j=0nbaj.Lemma 4.4.
The setℰ has cardinality (p-1)pepn-e+n-1.Proof.
Each elementg of the group P=B⋊Q has the form g=wmx(f) for a unique integer m∈{0,1,…,pn-1} and a unique function f∈ℱ. We will argue that g∈ℰ if and only if  x(f)∈kerφ while p does not divide m. From this it will follow that, to construct an element g∈ℰ, there are (p-1)pn-1 choices for m and |kerφ|=pepn-e choices for f.
BecauseP/B is cyclic of order pn, the condition 〈B,g〉=P holds if and only if  the coset gB=wmx(f)B=wmB has order pn as an element of P/B. Since the subgroups Q=〈w〉 and B intersect trivially, the coset wmB has order pn if and only if  the element wm has order pn. Recalling that the element w has order pn, we see that the element wm has order pn if and only if  p does not divide m. Therefore the condition 〈B,g〉=P holds if and only if  p does not divide m.
Also becauseP/B is cyclic of order pn, the condition 〈B,g〉=P implies that the order of the element g is divisible by pn. Henceforth we suppose that p does not divide m. To complete the proof, it suffices to show that gpn=1 if and only if  x(f)∈kerφ.
Writey=wm. Thus g=yx(f). Using Philip Hall's formula for raising the product of two elements to a power, along with the fact that ypn=1, we obtain
(4.7)gpn=∏j=0pn-1x(f)yj=∏j=0pn-1[∏u∈𝒰xuf(u)]yj=∏j=0pn-1[∏u∈𝒰(xuyj)f(u)]=∏u∈𝒰[∏j=0pn-1(xuyj)]f(u).
We define the elementz=x0x1⋯xpn-1∈B of order pe. Conjugation by w cyclically permutes the elements x0, x1, …, xpn-1. Since p does not divide m, conjugation by y=wm cyclically permutes the elements x0, x1, …, xpn-1 in some order. It follows that
(4.8)∏j=0pn-1(xuyj)=z.
From our work above, we deduce that
(4.9)gpn=∏u∈𝒰zf(u)=zs,wheres=∑u∈𝒰f(u)=φ(x(f)).
Recalling that the element z has order pe, we deduce that gpn=1 if and only if  x(f)∈kerφ.We will now complete the proof of Theorem A. SinceD⊆B while D and B are characteristic subgroups of P, every automorphism of P maps the set B-D to itself. Because xpn-1∈B and φ(xpn-1)=1, we have xpn-1∈B-D. Since B is a characteristic subgroup of P, every automorphism of P maps the set ℰ to itself. Note that w∈ℰ. Thus for each automorphism σ∈Aut(P), we have xpn-1σ∈B-D and wσ∈ℰ.Let𝒮 be the set consisting of all ordered pairs (a,b) such that a∈B-D and b∈ℰ. We now define the mapping Ψ:Aut(P)→𝒮 as follows. For each automorphism σ∈Aut(P) we let Ψ(σ)=(xpn-1σ,wσ). By the last sentence of the preceding paragraph, the mapping Ψ is well defined. Since {xpn-1,w} is a generating set for the group P, every automorphism of P is determined by where it maps the two elements xpn-1 and w, and so the mapping Ψ is injective. We now argue that |Aut(P)|=|𝒮|, an equality that would force the mapping Ψ to be a bijection, thereby completing the proof of Theorem A.Using|B:D|=p and |B|=pepn, we obtain |B-D|=(p-1)pepn-1. It is clear that |𝒮|=|B-D|·|ℰ|, and so by Lemma 4.4 we deduce that |𝒮|=(p-1)2p2epn+n-e-2. On the other hand, in the Introduction we calculated that |Aut(P)|=(p-1)2p2epn+n-e-2.



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*Source: 245617-2009-11-24.xml*