aa r X i v : . [ m a t h . G R ] F e b On factorizations of finite groups
Mikhail I. Kabenyuk
Abstract.
Let G be a finite group and let { A , . . . , A k } be a collection ofsubsets of G such that G = A . . . A k is the product of all the A i and card( G ) =card( A ) . . . card( A k ). We shall write G = A · . . . · A k , and call this a k -fold factorization of the form (card( A ) , . . . , card( A k )). We prove that for anyinteger k ≥ G of order n and a factorization of n = a . . . a k into k factors, with a i >
1, such that G has no k -fold factorizationof the form ( a , . . . , a k ). Keywords: factorization of a finite group, product of subsets, finite group,finite simple group, Sylow subgroup, centralizer involution
MSC:
Let G be a finite group and let { A , . . . , A k } be a collection of subsetsof G such that G = A . . . A k is the product of all the A i and card( G ) =card( A ) . . . card( A k ). We shall write G = A · . . . · A k , and call this a k -foldfactorization of the form (card( A ) , . . . , card( A k )).In [1] G. M. Bergman proved the following: Lemma 1 ([1, Lemma 2.1]). If G = A · . . . · A k is a factorization of afinite group G and A and A k both contain the identity element e of group G , then the order of the subgroup generated by A is a multiple of card( A ) ,and the order of the subgroup generated by A k is a multiple of card( A k ) . In the same article, the following question is formulated.
Question 2 ([1, Question 2.4]).
If a finite group G has a factorization G = A · A · A , must card( A ) divide the order of the subgroup H of G generated by A ?The answer to this question is negative. Let G be the alternatinggroup on 4 elements, a group of order 12. Let’s put A = { e, (14)(23) } , B = { e, (132) } , C = { e, (124) , (142) } . Composition of permutations willbe performed from left to right as it is accepted in [3]. It is easy to verifythat G = A · B · C . The order of the subgroup generated by B is 3 andcard( B ) = 2.The following statement is also proved in [1] (see also [5]). Proposition 3 ([1, Proposition 1.4]).
Let G be the alternating group n elements. Then G has no factorization A · B · C with card( A ) = 2 , card( B ) = 3 , card( C ) = 2 . This gives a negative answer to a question of M. H. Hooshmand [2,Problem 19.35] for k = 3. Problem 4.
Let G be a finite group of order n . a) Is it true that for every factorization n = a . . . a k there exist subsets A , . . . , A k such that card( A ) = a , . . . , card( A k ) = a k and G = A · . . . · A k ?b) The same question for the case k = 2 . If are considered only factorizations of n into factors >
1, then the prob-lem remains open for all k >
Theorem 5.
For any integer k ≥ there exist a finite group G oforder n and a factorization of n = a . . . a k into k factors, with a i > , suchthat G has no k -fold factorization of the form ( a , . . . , a k ) . Lemma 6 ([1, Lemma 2.2]).
If for some positive integers a , . . . , a k , G has a k -fold factorization with card( A i ) = a i ( i = 1 , . . . , k ) , then it has sucha factorization in which all A i contain e . Proposition 7.
Let G be a finite group with the following properties: (i) All involutions of G are conjugate. (ii) Each element of G has either order or an odd order.Then G has no -fold factorization A · B · C with card( A ) = 2 = card( C ) and card( B ) = card( G ) / .Proof. We argue by contradiction. Suppose G = A · B · C is afactorization with card( A ) = 2 = card( C ) and card( B ) = card( G ) /
4. Itfollows from Lemma 6 that we can assume without loss of generality that e ∈ A and e ∈ C . Then by Lemma 1 the subgroups generated by A and C are cyclic subgroups of even orders. By the condition (ii) it follows thatboth these subgroups have order 2, so A and C are subgroups of G .Recall that the set of elements AxC , where x is some fixed element of G ,is called a double coset. Two double cosets AxC and
AyC are either disjointor identical [3, p. 14]. Since the group G is a disjoint union of double cosetsof the form AxC and G = A · B · C , then the number of double cosets in G is card( B ) and card( AxC ) = 4 for any x ∈ G .By the condition (i) there exists x ∈ G that x − Ax = C . Then AxC = x ( x − Ax ) C = xC = xC. So, card(
AxC ) = card( xC ) = 2 <
4. This is a contradiction; the proof ofthe proposition is complete.
Corollary 8.
Let G be a finite group of order n satisfying the conditionsof Proposition 7 and let Ω( n ) be the number of prime factors of n (with ultiplicity). Then for any integer k in the interval ≤ k ≤ Ω( n ) and forany integers a , . . . , a k such that n = a . . . a k and a = a k = 2 the group G has no k -fold factorization A · . . . · A k with card( A i ) = a i ( i = 1 , . . . , k ) .Proof. Suppose that for some k the group G has k -fold factorization G = A · . . . · A k with card( A ) = card( A k ) = 2. Then G = A · B · A k ,where B = A · . . . · A k − ; a contradiction with Proposition 7. Proof of Theorem 5.
Let q = p s be a prime power with s ≥ GF ( q ) be Galois field of order q and let GF ( q ) ∗ be the multiplicative groupof the field GF ( q ) Consider a group G ( q ) and its the Sylow p -subgroup P : G ( q ) = (cid:18) GF ( q ) ∗ GF ( q )0 1 (cid:19) , P = (cid:18) GF ( q )0 1 (cid:19) . It is easy to see that P is a normal subgroup of G ( q ) and an elementaryAbelian p -group. Observe that any two non-trivial elements of P are conju-gate in G . Indeed, if u, v ∈ GF ( q ), uv = 0 and a = v/u then (cid:18) a
00 1 (cid:19) (cid:18) u (cid:19) (cid:18) a
00 1 (cid:19) − = (cid:18) au (cid:19) = (cid:18) v (cid:19) . If g ∈ G ( q ) and g / ∈ P , then g k = (cid:18) a u (cid:19) k = (cid:18) a k (1 + a + . . . + a k − ) u (cid:19) = a k a k − a − u ! , where a = 1. It follows that the elements a ∈ GF ( q ) and g ∈ G ( q ) are ofthe same order. Hence, if g / ∈ P , then the order of g is relatively prime to p . It is clear also that if g ∈ P and g = e , then the order of g is p .If q = 2 s , then all the assumptions of Corollary 8 hold. Since n =card( G ( q )) = 2 s (2 s −
1) it follows that Ω( n ) ≥ s + 1 Therefore, for anyinteger k in the interval 3 ≤ k ≤ s + 1 and for any integers a , . . . , a k suchthat n = a . . . a k and a = a k = 2 the group G ( q ) has no k -fold factorization A · . . . · A k with card( A i ) = a i ( i = 1 , . . . , k ). Since the choice of s is at ourdisposal, Theorem 5 is proved. Note on simple groups.
It is clear that the condition (ii) of Proposi-tion 7 is equivalent to(ii’) The centralizer of every involution of G is an elementary Abeliangroup.The alternating group A satisfies the conditions (i) and (ii’). This meansthat A has no 3-fold factorization A · B · C with card( A ) = 2 = card( C )3nd card( B ) = 15. Moreover, A has no 4-fold factorizations of the forms(2 · · ·
2) and (2 · · · SL (2 , q ), where q = 2 s , s ≥ SL (2 , q ) is 2 s (2 s − SL (2 , q ) is anelementary Abelian group of order 2 s . Let P consists of all the unitriangularmatrices of the form (cid:18) u (cid:19) , u ∈ GF ( q ) . Then P is the Sylow 2-subgroup of G . It is easy to see that SL (2 , q ) satisfiesthe conditions (i) and (ii’). Therefore, SL (2 , q ) with q = 2 s , s ≥
2, has no3-fold factorization A · B · C with card( A ) = 2 = card( C ) and card( B ) =2 s − (2 s − G satisfiesthe conditions (i) and (ii’), then G is isomorphic to A or SL (2 , q ), where q = 2 s and s ≥ Proposition 9.
Let p be prime and let G be a finite group with thefollowing properties: (i) All elements of order p are conjugate in G . (ii) Each element of G has either order p or an order relatively primeto p .Then G has no -fold factorization A · B · C such that A and C are thesubgroups of G and card( A ) = p = card( C ) and card( B ) = card( G ) /p . References [1] G. M. Bergman,
A note on factorizations of finite groups , Journal of theIranian Mathematical Society, Vol. 1 No. 2 (2020), pp. 157 – 161.[2] E.I. Khukhro and V.D. Mazurov (eds.),
Unsolved Problems in GroupTheory, The Kourovka Notebook , 19, Sobolev Institute of Math-ematics, Novosibirsk (2018); arXiv:1401.0300 [math.GR], see alsohttps://kourovka-notebook.org/ for the current updates.[3] M. Hall,
The theory of groups , New York, 1959.[4] D.Gorenstein,
Finite groups in which Sylow -subgroups are abelian andcentralizers of involutions are solvable , Canad. J. Math. 17 (1965), 860– 906. 45] https://mathoverflow.net/questions/316262/is-each-finite-group-multifactorizableKemerovo State University, Kemerovo, Russia E-mail address: kk