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Automorphisms of Regular Wreath Product -Groups
We present a useful new characterization of the automorphisms of the regular wreath product group of a finite cyclic -group by a finite cyclic -group, for any prime , 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(), where is the regular wreath product of a finite cyclic -group by an arbitrary finite -group.
Let denote the regular wreath product group where is an arbitrary nontrivial finite -group, for some prime , and where is an any finite cyclic -group. Thus is the semidirect product , where is a direct product of copies of , and where acts via automorphisms on by regularly permuting these direct factors.
In , Houghton determines some information on the structure of the automorphism group . Using this work of Houghton (see also [2, Chapter 5]), it is possible to calculate the order of . Our first result in this paper is to present an alternative method for calculating the order of . Our approach to this calculation is to apply the Automorphism Counting Formula (established in ), a general formula for the order of the automorphism group of a monolithic finite group in terms of information about the complex characters of and information about how 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 -group is monolithic if and only if its center is cyclic. Let and . Throughout this paper we assume that , which excludes only the case where and , for which is dihedral of order .
Theorem 1.1. has order , where .
Because the dihedral group of order has an automorphism group of order , the condition 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 be any prime-power larger than such that is the full -part of . Then the general linear group has exactly one conjugacy class of subgroups whose members are isomorphic to .
Now suppose that the group of order is cyclic. Since has order , Theorem 1.1 yields . Using knowledge of and little more than an elementary counting argument, we obtain a useful new characterization of the automorphisms of . Before stating this characterization, we establish some notation.
Hypothesis 1.3. Assume that the group of order is cyclic. Let be a collection of elements of order that constitutes a generating set for the homocyclic group of exponent and of rank . Let be a generator for the cyclic group and suppose that for each and that .
Under Hypothesis 1.3, it is clear that is a generating set for the group , and so every automorphism of is determined by where it maps these two elements.
Neumann  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 is always a characteristic subgroup of for the particular class of wreath product groups considered in this paper. Nevertheless, in our proof of Theorem 1.1 we present our own brief argument (see Step 7) that is a characteristic subgroup of . From this fact it follows that is a characteristic subgroup of .
We are now ready to state the main result of this paper.
Theorem 1 A. Assume Hypothesis 1.3. Then the group is cyclic of order , and therefore has a unique maximal subgroup which one denotes as , and so is a characteristic subgroup of that satisfies . Let denote the set of all elements of order that satisfy the condition . Then for each pair of elements such that and , there exists an automorphism of that maps to and maps to . Furthermore, every automorphism of is of this type.
In the notation of Theorem A, the information that we have about the subgroup and the set makes it clear that every automorphism of maps the set to itself and maps the set to itself. It is not difficult to see that the element belongs to the set and that the element belongs to the set . From this perspective, we might summarize Theorem A as stating that every mapping that could possibly be an automorphism of actually is an automorphism of .
Theorem A gives us a factorization of , namely, with , where . Houghton's main result in  is a factorization of , namely, with , where denotes the group of inner automorphisms of induced by elements of , and where is the image of the usual embedding of in (see ). In particular . Since , these two factorizations are the same if and only if . However, permutes the elements with lying in a regular orbit, and so . Hence these two factorizations are the same if and only if , which happens only when .
We now discuss an application of Theorem A. In  we classify up to isomorphism the nonabelian subgroups of the wreath product group for an arbitrary prime and positive integer such that . In  we use the characterization of the elements of that is provided by Theorem A to compute the index for each group of class or larger appearing in this classification. For each such group , we then observe that this index is equal to the order of the automorphism group , from which we deduce that the group is isomorphic to , which says that the full automorphism group is realized inside the group .
Let denote the set of irreducible ordinary characters of a finite group .
For each finite group and prime-power , let denote the smallest positive integer such that the general linear group contains a subgroup that is isomorphic to . Thus is the minimal degree among all the faithful -representations of the group , where denotes the field with elements. For any groups and such that , we have .
Definition 2.1. Let be a monolithic finite group, let be a prime-power that is relatively prime to the order of , and let . We say that the ordered triple is a monolithic triple in case every faithful irreducible ordinary character of has degree at least . Assuming that is a monolithic triple, we define to be the set of all faithful irreducible ordinary characters of of degree . We say that the monolithic triple is good provided that every value of each character belonging to the set is a -linear combination of complex st roots of unity.
Theorem 2.2. Let be a good monolithic triple. Suppose that has a unique conjugacy class of subgroups whose members are isomorphic to . Let be any subgroup of that is isomorphic to . Then .
In our proof of Theorem 1.1, the idea is to define a good monolithic triple with that satisfies the hypothesis of Theorem 2.2. The conclusion of Theorem 2.2 would then yield provided that we know in advance and .
Given a monolithic group , in order to define a good monolithic triple we must choose an appropriate prime-power and then calculate . The following result may be used to calculate for certain groups and prime-powers .
Lemma 2.3. Let be any finite group containing an abelian -subgroup of exponent and of rank , where is a prime. Let be any field containing a primitive th root of unity. If there exists a faithful -representation of of degree , then .
Proof. The hypotheses yield . It remains to show that . The hypothesis on implies that every irreducible -representation of has degree and that the characteristic of the field is not . Let be any faithful -representation of , and let be its degree. By Maschke's theorem, is similar to a faithful -representation consisting of diagonal matrices. Let be the subgroup of consisting of all diagonal matrices of order dividing . Then while is homocyclic of exponent and of rank . Since is faithful, indeed is an abelian -group of rank . It follows that . Therefore , as desired.
One of the hypotheses of Theorem 2.2 is that the general linear group has a unique conjugacy class of subgroups whose members are isomorphic to . The following result (Lemma in ) is useful for establishing this condition in certain situations.
Lemma 2.4. Let be a field containing a primitive th root of unity, where is some prime and is some positive integer. Let be any finite group containing an abelian normal -subgroup of exponent and of rank . Then every faithful -representation of of degree is similar to a representation such that consists of diagonal matrices and consists of monomial matrices.
Using Theorem 2.2 to calculate the order of the automorphism group for a given monolithic triple requires that we know in advance the cardinality of the set that was defined in Definition 2.1. The following result is helpful for calculating the cardinality of the set in certain situations.
Lemma 2.5. Let be a prime and let be a monolithic finite -group. One defines the set . Let be a nonnegative integer and suppose that every character belonging to the set has degree . Then .
Proof. We define the set . Let be the unique minimal normal subgroup of , and note that . Hence the set may be identified with the set . We have , and so . By Corollary in , along with the fact that is a disjoint union, we deduce that Solving this equation for , we obtain the desired conclusion.
Using Theorem 2.2 to calculate the order of the automorphism group for a given monolithic triple requires that we know in advance the order of the normalizer of a certain subgroup in the general linear group . The following result (which is part of Theorem in ) is useful for this task in certain situations.
Theorem 2.6. Let where is any prime-power and is any positive integer. Let be the field with elements, let be any nontrivial subgroup of the multiplicative group , and let be the group of all diagonal matrices in having the property that each entry along the diagonal belongs to . Let be the subgroup of consisting of all permutation matrices, and note that . Let be any transitive subgroup of the symmetric group and let . If is a characteristic subgroup of , then .
The following rather specialized result will be used in our proof of Theorem 1.1.
Lemma 2.7. Let be any prime and let , , and be positive integers such that . Then the condition holds if and only if and .
Proof. First, an easy inductive argument shows that whenever . Now suppose that holds. First we show that . Assuming instead that , we get , forcing , a contradiction. Hence , and so , which forces each of the positive integers , , and to be . Therefore and , as desired. The reverse implication is trivial.
The next two results on permutation groups will be used later in this article.
Lemma 2.8. Let and be isomorphic transitive subgroups of order of the symmetric group . Then and are conjugate subgroups of .
Proof. For each and each , let denote the image of under . For , the maps defined by are bijections. Let be an isomorphism. The composition is an element of . It suffices to show that for each . A straightforward calculation (left to the reader) yields for arbitrary .
Theorem 2.9. Let be any transitive subgroup of order in the symmetric group . Then the normalizier is isomorphic to the holomorph .
The following basic lemma is needed for our proof of Theorem 2.9.
Lemma 2.10. Let be a group of permutations of a set , let be a transitive subgroup of , and let . For each , the stabilizer subgroup is trivial.
Proof. Let . To prove that , it suffices to show that for arbitrary , since acts faithfully. There exists such that . Since , we have , and so .
Proof of Theorem 2.9. Let be a group that is isomorphic to . Let where . First we identify a subgroup of that is isomorphic to and that centralizes . The rule defines an injective homomorphism , where is the inner automorphism induced by . Let . For , , observe that
Next we embed as a subgroup of in such a way that becomes a transitive (in fact regular) subgroup of . Since , the action of on the set consisting of the right cosets of in is faithful. We now argue that the action of on is regular. Since , it suffices to show that each nonidentity element of fixes no element of . Let and such that fixes . Thus and so . Since , we obtain , and so , as desired. Now label the members of as the numbers . In this way we regard as a subgroup of .
Since and are isomorphic transitive subgroups of order in , by Lemma 2.8 we may complete the proof by showing that . Write . Lemma 2.10 implies that every orbit in the action of on has size . Hence divides . But since centralizes , we have . It follows that .
Write . By the -Mod- Theorem, the integer divides , which says that divides . Recalling that , this says that divides . But since , we have . It follows that .
3. Proof of Theorem 1.1
Let be a collection of elements of order that constitutes a generating set for the homocyclic group of exponent and of rank . We now define an action of the group on the set . For each pair , , we let , where the product is computed in . This action naturally gives rise to an action of via automorphisms on the group . Let denote the semidirect product group corresponding to this action. Let denote the set consisting of all functions from into the additive group . For each function , we define the element Each element of has the form for some unique function . We define the element of order by letting denote the product of all the elements for .
Step 1. For each subgroup of , the centralizer is equal to the set of all elements such that the function is constant on each of the left cosets of in .
Proof. Let be a transversal for the left cosets of in . For each , observe that the set is an orbit in the action of on the set of generators for .
Step 2. The group is monolithic, and its center is the cyclic group of order .
Proof. Since is abelian and the action of via automorphisms on is faithful, the center of is . By Step 1, is the cyclic group generated by the element . Finally, since is a -group whose center is cyclic, is indeed monolithic.
Following standard notation (see ), we define the inertia subgroup of any character as the subgroup .
Step 3. For each character such that , every irreducible constituent of the induced character is not faithful.
Proof. For each pair of functions , we define the dot product to be the value
Let be any primitive complex th root of unity. For each function , we define the character by for every function . It is clear that every irreducible ordinary character of is of the form for some function .
Let such that . Since is equal to the intersection of the kernels of the irreducible constituents of , it suffices to show that . Because , we have for some nontrivial subgroup of . Let be any transversal for the left cosets of in . Since , the prime divides . Since , we have for some function . Because the character is -invariant, the function must be constant on each left coset of in . This says that for each , there exists a value such that for each element .
By Step 2, is the unique minimal normal subgroup of . Note that for the constant function defined as for . Observe that For each , using the fact that is divisible by , we deduce that It follows that , which yields . Hence . Using and , we obtain , as desired.
We define the set .
Step 4. For each character we have , and for each element the value is a sum of complex th roots of unity. Furthermore .
Proof. Let be arbitrary and let be any irreducible constituent of the restriction . Hence is an irreducible constituent of the induced character . Since and since is faithful, Step 3 yields . By the Clifford Correspondence [7, Theorem 6.11], it follows that is irreducible, and so . Since while is abelian, we have . Therefore .
Since with and , the character vanishes off . Furthermore, because is an abelian -group of exponent , every value of is a complex th root of unity. By Theorem in , the restriction is a sum of conjugates of in . Hence for each element , the value is a sum of complex th roots of unity.
Finally, Lemma 2.5 yields , as desired.
Let be any prime-power such that is the full -part of . Let where is the field with elements. Let , , and denote the subgroups of consisting of all diagonal matrices, permutation matrices, and monomial matrices, respectively. Note that and that is isomorphic to the symmetric group of degree . Let denote the subgroup of consisting of all diagonal matrices of order dividing . Thus is homocyclic of exponent and of rank . Note that is the unique Sylow -subgroup of the abelian group , and that is a separator subgroup of .
We will now define a faithful representation . Recall that is a collection of elements of order that constitutes a generating set for the homocyclic group of exponent and of rank . We index the rows and the columns of the matrices in by the elements of the group . We choose an arbitrary element of order in the cyclic multiplicative group of nonzero elements in the field . For each , we define to be the diagonal matrix in whose -entry is , and each of whose other diagonal entries is . Thus consists of diagonal matrices. We define to be the right regular representation of the group . Thus consists of permutation matrices and is a regular subgroup of the symmetric group . The action of by conjugation on inside the group is similar to the action of by conjugation on inside the group . Thus, since and , we have a faithful representation whose image is a subgroup of .
Step 5. .
Proof. Recall that is a faithful -representation of of degree ; use Lemma 2.3.
The next step establishes Theorem 1.2.
Step 6. Every faithful -representation of of degree is similar to .
Proof. By Lemma 2.4, every faithful -representation of of degree is similar to a faithful -representation such that and . Since is the unique Sylow -subgroup of , indeed . Since is faithful, the -groups and are homocyclic of exponent and of rank . It follows that . That is the unique Sylow -subgroup of yields . Satz II.7.2(a) in  yields , so . Let be a Sylow -subgroup of . Thus is a Sylow -subgroup of . Since is a -subgroup of , Sylow's theorem asserts that is similar (by a matrix in ) to a representation such that . We have , since . Thus and are regular subgroups of the symmetric group , and are both isomorphic to . By Lemma 2.8, conjugation by some element of maps to . Conjugation by the unique preimage of this element under the natural isomorphism maps to . Hence is similar to .
Step 7. is a characteristic subgroup of .
Proof. We argue that is the only abelian normal subgroup of index in . Let be an abelian normal subgroup of such that and . Write with and let . We now argue that . Since while , we have . Because , Dedekind's lemma yields , and so . From this we obtain . Since and are abelian, we have . It follows that and . By Step 1, we have . Since , it follows that . Thus . By Lemma 2.7, this contradicts the hypothesis .
Step 8. The normalizer has order .
Proof. Using and , we obtain . By Step 7 and the fact that is faithful, is a characteristic subgroup of . Since is a regular subgroup of the symmetric group and since , Theorem 2.9 implies that the normalizer is isomorphic to the holomorph of . Therefore . The statement now follows from Theorem 2.6.
Step 9. .
Proof. By Steps 2, 4, and 5, is a good monolithic triple and . Thus Step 4 yields . By Step 6, belongs to the unique conjugacy class of subgroups of whose members are isomorphic to . In view of Step 8, Theorem 2.2 yields where .
4. Proof of Theorem A
Assume Hypothesis 1.3. Let denote the set of all functions from the set into the additive group . For each function , we define the element Each element of has the form for some unique . The mapping defined by is a surjective homomorphism. Hence is cyclic of order . To establish Theorem A, our first task is to prove that is cyclic of order . For this it suffices to show that .
Lemma 4.1. For each function , the commutator element has the form
Proof. Note that . Conjugating by , we obtain Since , the result follows.
Theorem 4.2. .
Proof. Let denote the subgroup of that is generated by all elements of the form with . Using , we can show that . Since while is abelian, it is clear that . Hence it suffices to show that .
To show that , we must verify that for each , but this is obvious by Lemma 4.1. Next we argue that . An arbitrary element of has the form for some function satisfying . To establish that , we will now define a particular function such that . Let , and for each let . It follows that for each we have . Furthermore, using the condition , we obtain By Lemma 4.1, we deduce that . Therefore , as desired.
The cyclic group has a unique subgroup of index , namely, . Let be the group consisting of all those elements in such that . It is clear that and .
Corollary 4.3. and are characteristic subgroups of .
Proof. By Step 7 in the proof of Theorem 1.1, is a characteristic subgroup of . It follows that is a characteristic subgroup of . By Theorem 4.2, we deduce that is a characteristic subgroup of . Since is cyclic, is the only subgroup of that satisfies the conditions and . Because and are characteristic subgroups of , it follows that is a characteristic subgroup of .
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 and any elements and belonging to some group, the element may be written as This says that . Furthermore, in case all the conjugates of by powers of commute with each other (which is automatically true if is contained in an abelian normal subgroup of any group containing and ), this formula becomes
Lemma 4.4. The set has cardinality .
Proof. Each element of the group has the form for a unique integer and a unique function . We will argue that if and only if while does not divide . From this it will follow that, to construct an element , there are choices for and choices for .
Because is cyclic of order , the condition holds if and only if the coset has order as an element of . Since the subgroups and intersect trivially, the coset has order if and only if the element has order . Recalling that the element has order , we see that the element has order if and only if does not divide . Therefore the condition holds if and only if does not divide .
Also because is cyclic of order , the condition implies that the order of the element is divisible by . Henceforth we suppose that does not divide . To complete the proof, it suffices to show that if and only if .
Write . Thus . Using Philip Hall's formula for raising the product of two elements to a power, along with the fact that , we obtain
We define the element of order . Conjugation by cyclically permutes the elements , , , . Since does not divide , conjugation by cyclically permutes the elements , , , in some order. It follows that From our work above, we deduce that Recalling that the element has order , we deduce that if and only if .
We will now complete the proof of Theorem A. Since while and are characteristic subgroups of , every automorphism of maps the set to itself. Because and , we have . Since is a characteristic subgroup of , every automorphism of maps the set to itself. Note that . Thus for each automorphism , we have and .
Let be the set consisting of all ordered pairs such that and . We now define the mapping as follows. For each automorphism we let . By the last sentence of the preceding paragraph, the mapping is well defined. Since is a generating set for the group , every automorphism of is determined by where it maps the two elements and , and so the mapping is injective. We now argue that , an equality that would force the mapping to be a bijection, thereby completing the proof of Theorem A.
Using and , we obtain . It is clear that , and so by Lemma 4.4 we deduce that . On the other hand, in the Introduction we calculated that .
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Copyright © 2009 Jeffrey M. Riedl. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.