Automorphisms of Regular Wreath Product p-Groups

We present a useful new characterization of the automorphisms of the regular wreath product group P 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.


Introduction
Let P 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| p e and |Q| p n .Throughout this paper we assume that p en ≥ 3, which excludes only the case where p 2 and e n 1, for which P is dihedral of order 8.Because the dihedral group of order 8 has an automorphism group of order 8, the condition p en ≥ 3 is a necessary hypothesis for Theorem 1.1.
The next result is a step along the way to proving Theorem 1.1.We mention it here.
Theorem 1.2.Let q be any prime-power larger than 1 such that p e is the full p-part of q − 1.Then the general linear group GL p n , q has exactly one conjugacy class of subgroups whose members are isomorphic to P .Now suppose that the group Q of order p n is cyclic.Since Aut Q has order p − 1 p n−1 , Theorem 1.1 yields |Aut P | p − 1 2 p 2ep n 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 group Q of order p n is cyclic.Let x 0 , x 1 , . . ., x p n −1 be a collection of elements of order p e that constitutes a generating set for the homocyclic group B of exponent p e and of rank p n .Let w be a generator for the cyclic group Q and suppose that x w u x u−1 for each u ∈ {1, . . ., p n − 1} and that x w 0 x p n −1 .
Under Hypothesis 1.3, it is clear that {x p n −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 A. Assume Hypothesis 1.3.Then the group B/ B, P is cyclic of order p e , 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 p n 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 x p n −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 subgroup D 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 x p n −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 of A Aut P , namely, 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 ⊆ C A x , these two factorizations are the same if and only if Q * ⊆ C A x .However, Q * permutes the elements x 0 , x 1 , . . ., x p n −1 with x x p n −1 lying in a regular orbit, and so Q * ∩ C A 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 Z p e Z p for an arbitrary prime p and positive integer e such that p e ≥ 3.In 6 we use the characterization of the elements of A Aut P that is provided by Theorem A to compute the index |N A H : C A 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 N A H /C A H is isomorphic to Aut H , which says that the full automorphism group Aut H is realized inside the group A Aut P .
In Section 3 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.
Let Irr G denote the set of irreducible ordinary characters of a finite group G.

Preliminaries
For each finite group G 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.Let G 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 In our proof of Theorem 1.1, 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 |N Γ H |.
Given a monolithic group G, 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.Let G be any finite group containing an abelian p-subgroup B of exponent p e and of rank r, where p is a prime.Let F be any field containing a primitive p e th root of unity.If there exists a faithful F-representation of G of degree r, then mindeg G, F r.
Proof.The hypotheses yield mindeg 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 Frepresentation of B, and let n be its degree.By Maschke's theorem, X is similar to a faithful Frepresentation Y consisting of diagonal matrices.Let E be the subgroup of GL n, F consisting of all diagonal matrices of order dividing p e .Then Y B ⊆ E while E is homocyclic of exponent p e 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 Theorem 2.2 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.Let F be a field containing a primitive p e th 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 p e 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 Theorem 2.2 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.Using Theorem 2.2 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.Let Γ 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 F 0 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 F 0 .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

International Journal of Mathematics and Mathematical Sciences 5
The following rather specialized result will be used in our proof of Theorem 1.1.Lemma 2.7.Let p be any prime and let e, n, and j be positive integers such that j ≤ n.Then the condition ep n−j p j − 1 ≤ j holds if and only if p 2 and e n j 1.
Proof.First, an easy inductive argument shows that 2 j − 1 > j whenever j ≥ 2. Now suppose that ep n−j p j − 1 ≤ j holds.First we show that j 1.Assuming instead that j ≥ 2, we get p j −1 ≥ 2 j −1 > j, forcing ep n−j p j −1 > j, a contradiction.Hence j 1, and so ep n−1 p 1 −1 ≤ 1, which forces each of the positive integers e, p n−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.The following basic lemma is needed for our proof of Theorem 2.9.Lemma 2.10.Let G be a group of permutations of a set Ω, let H be a transitive subgroup of G, and let C C G H .For each α ∈ Ω, the stabilizer subgroup C α is trivial.
Proof.Let x ∈ C α .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 Proof of Theorem 2.9.Let G 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 → ϕ x x −1 defines an injective homomorphism θ : G → V , where ϕ x ∈ A is the inner automorphism induced by x.Let D θ G .For x, y ∈ G, observe that Next we embed V as a subgroup of S in such a way that G becomes a transitive in fact regular subgroup of S. Since core V 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.

Proof of Theorem 1.1
Let {x u | u ∈ Q} be a collection of elements of order p e that constitutes a generating set for the homocyclic group B of exponent p e and of rank |Q| p n .We now define an action of the group Q on the set {x u | u ∈ Q}.For each pair u, v ∈ Q, we let x v u x uv , 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 Z p e .For each function f ∈ F, we define the element 3.1 Each element of B has the form x f for some unique function f ∈ F. We define the element z ∈ B of order p e by letting z denote the product of all the elements x u for u ∈ Q.
Step 1.For each subgroup L of Q, the centralizer C B 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.Let T be a transversal for the left cosets of L in Q.For each t ∈ T , observe that the set {x u | u ∈ tL} is an orbit in the action of L on the set of generators {x u | u ∈ Q} for B.
Step 2. The group P is monolithic, and its center is the cyclic group z of order p e .
Proof.Since B is abelian and the action of Q via automorphisms on B is faithful, the center of Step 1, C B 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 I P θ {x ∈ P | θ x θ}.
Step 3.For each character θ ∈ Irr B such that I P θ > B, every irreducible constituent of the induced character θ P is not faithful.
Proof.For each pair of functions f, g ∈ F we define the dot product f • g to be the value

3.2
Let be any primitive complex p e th root of unity.For each function g ∈ F, we define the character ϕ g ∈ Irr B by ϕ g x f f•g for every function f ∈ F. It is clear that every irreducible ordinary character of B is of the form ϕ g for some function g ∈ F.
Let θ ∈ Irr B such that I P θ > B. Since ker θ P is equal to the intersection of the kernels of the irreducible constituents of θ P , it suffices to show that ker θ P > 1.Because P B Q, we have I P 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 ϕ g 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 c t ∈ Z p e such that g u c t for each element u ∈ tL.By Step 2, z p e−1 is the unique minimal normal subgroup of P .Note that z p e−1 x f for the constant function f ∈ F defined as f u p e−1 for u ∈ Q. Observe that f u g u .

3.3
For each t ∈ T , using the fact that |tL| |L| is divisible by p, we deduce that u∈tL f u g u u∈tL p e−1 c t |L|p e−1 c t 0.

3.4
It follows that f • g 0, which yields z p e−1 x f ∈ ker θ.Hence z p e−1 ⊆ ker θ.Using ker θ P core P ker θ and 1 < z p e−1 P , we obtain 1 < z p e−1 ⊆ ker θ P , as desired.
We define the set A {ψ ∈ Irr P | ψ is faithful}.
Step 4. For each character χ ∈ A we have χ 1 p n , and for each element x ∈ P the value χ x is a sum of complex p e th roots of unity.Furthermore |A| p − 1 |P |/p 2n 1 .
Proof.Let χ ∈ A be arbitrary and let θ ∈ Irr B be any irreducible constituent of the restriction χ B .Hence χ is an irreducible constituent of the induced character θ P .Since B ⊆ I P θ and since χ is faithful, Step 3 yields I P θ B. By the Clifford Correspondence 7, Theorem 6.11 , it follows that θ P is irreducible, and so χ θ P .Since θ ∈ Irr B while B is abelian, we have θ 1 1. Therefore χ 1 θ P 1 |P : B|θ 1 |Q| p n .Since χ θ P with θ ∈ Irr B and B P , the character χ vanishes off B. Furthermore, because B is an abelian p-group of exponent p e , every value of θ is a complex p e th root of unity.By Theorem 6.2 in 7 , the restriction χ B is a sum of conjugates of θ in P .Hence for each element x ∈ B, the value χ x is a sum of complex p e th roots of unity.
Let q > 1 be any prime-power such that p e is the full p-part of q − 1.Let Γ GL p n , 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 p n .Let E denote the subgroup of Γ consisting of all diagonal matrices of order dividing p e .Thus E is homocyclic of exponent p e and of rank p n .Note that E is the unique Sylow p-subgroup of the abelian group D, and that E is a separator subgroup of Γ.

International Journal of Mathematics and Mathematical Sciences
We will now define a faithful representation Z : P → Γ. Recall that {x u | u ∈ Q} is a collection of elements of order p e that constitutes a generating set for the homocyclic group B of exponent p e and of rank |Q| p n .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 p e in the cyclic multiplicative group of nonzero elements in the field F. For each u ∈ Q, we define Z x u 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 Step Proof.Using P 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 that for each u ∈ {0, 1, . . ., p n − 2} we have f u 1 − f u g u .Furthermore, using the condition g 0 g 1 • • • g p n − 1 0, we obtain By Lemma 4.1, we deduce that x f , w x g .Therefore x g ∈ B, w , as desired.
The cyclic group Z p e has a unique subgroup of index p, namely, pZ p e {pa | a ∈ Z p e }.Let D be the group consisting of all those elements x f in B such that ϕ x f ∈ pZ p e .It is clear that ker ϕ ⊆ D ⊆ B and |B : D| 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 integer n and any elements a and b belonging to some group, the element ab n may be written as Proof.Each element g of the group P B Q has the form g w m x f for a unique integer m ∈ {0, 1, . . ., p n − 1} and a unique function f ∈ F. We will argue that g ∈ E if and only if x f ∈ ker ϕ while p does not divide m.From this it will follow that, to construct an element g ∈ E, there are p − 1 p n−1 choices for m and |ker ϕ| p ep n −e choices for f.Because P/B is cyclic of order p n , the condition B, g P holds if and only if the coset gB w m x f B w m B has order p n as an element of P/B.Since the subgroups Q w and B intersect trivially, the coset w m B has order p n if and only if the element w m has order p n .Recalling that the element w has order p n , we see that the element w m has order p n 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.

Lemma 2 . 5 . 1 2 ψ∈B ψ 1 2
Let p be a prime and let P be a monolithic finite p-group.One defines the set A {ψ ∈ Irr P | ψ is faithful}.Let n be a nonnegative integer and suppose that every character belonging to the set A has degree p n .Then |A| |P | p − 1 /p 2n 1 .Proof.We define the set B Irr P − A. Let N be the unique minimal normal subgroup of P , and note that B {ψ ∈ Irr P | N ⊆ ker ψ}.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 |P | ψ∈A ψ |A|p 2n |P | p .2.1 Solving this equation for |A|, we obtain the desired conclusion.

Since
H and G are isomorphic transitive subgroups of order n in S, by Lemma 2.8 we may complete the proof by showing that N S G V .Write C C S 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. Write N N S 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.

Corollary 4 . 3 .
ker ϕ and D are characteristic subgroups of P .Proof.By Step 7 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 ker ϕ is a characteristic subgroup of P .Since B/ker ϕ is cyclic, D is the only subgroup of P that satisfies the conditions ker ϕ ⊆ D ⊆ B and |B : D| p.Because B and ker ϕ are characteristic subgroups of P , it follows that D is a characteristic subgroup of P .

Lemma 4 . 4 .
a n a − n−1 ba n−1 a − n−2 ba n−2 • • • a −2 ba 2 a −1 ba 1 b.4.5This says that ab n a n b a n−1 b a n−2 • • • b a 2 b a 1 b.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 becomesab n a n bb a 1 b a 2 • • • b a n−2 b a n−1The set E has cardinality p − 1 p ep n −e n−1 .
Lemma 2.8.Let H 1 and H 2 be isomorphic transitive subgroups of order n of the symmetric group Sym n .Then H 1 and H 2 are conjugate subgroups of Sym n .Proof.For each α ∈ Ω {1, . . ., n} and each x ∈ Sym n , let α • x denote the image of α under x.For i ∈ {1, 2}, the maps f i : H i → Ω defined by f i x 1 • x are bijections.Let θ : H 1 → H 2 be an isomorphism.The composition y f 2 θf −1 1 : Ω → Ω is an element of Sym n .It suffices to show that y −1 xy θ x for each x ∈ H 1 .A straightforward calculation left to the reader yields α • y −1 xy α • θ x for arbitrary α ∈ Ω.Let H be any transitive subgroup of order n in the symmetric group S Sym n .Then the normalizier N S H is isomorphic to the holomorph H Aut H .
5. mindeg P, F p n .Proof.Recall that Z is a faithful F-representation of P of degree p n ; use Lemma 2.3.Every faithful F-representation of P of degree p n is similar to Z.Proof.By Lemma 2.4, every faithful F-representation of P of degree p n is similar to a faithful F-representation X such that X B ⊆ D and X P ⊆ M. Since E is the unique Sylow psubgroup of D, indeed X B ⊆ E. Since X is faithful, the p-groups X B and E are homocyclic of exponent p e and of rank p 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 p n , 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 that B is the only abelian normal subgroup of index p n in P .Let A be an abelian normal subgroup of P such that |P : A| p n and A / B. Write |AB : B| p j 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| p j .Since A and B are abelian, we have A ∩ B ⊆ Z AB .It follows that A ∩ B ⊆ C B L ⊆ B and |B : C B L | ≤ |B : A ∩ B| p j .By Step 1, we have |C B L | p e |Q:L| p ep n−j .Since |B| p ep n , it follows that |B : C B L | p ep n−j p j −1 .Thus ep n−j p j − 1 ≤ j.By Lemma 2.7, this contradicts the hypothesis p en ≥ 3.
n .It follows that X B E. That E is the unique Sylow psubgroup of D yields E N Γ D .Satz II.7.2 a in 8 yields N Γ D M, so E M DS.