aa r X i v : . [ m a t h . G R ] A ug A Strict Inequality for a Minimal Degree of aDirect Product
Neil SaundersNovember 18, 2018
Abstract
The minimal faithful permutation degree µ ( G ) of a finite group G isthe least non-negative integer n such that G embeds in the symmetricgroup Sym ( n ). Work of Johnson and Wright in the 1970’s establishedconditions for when µ ( H × K ) = µ ( H )+ µ ( K ), for finite groups H and K . Wright asked whether this is true for all finite groups. A counter-example of degree 15 was provided by the referee and was added asan addendum in Wright’s paper. Here we provide a counter-exampleof degree 12. The minimal faithful permutation degree µ ( G ) of a finite group G is theleast non-negative integer n such that G embeds in the symmetric group Sym ( n ). It is well known that µ ( G ) is the smallest value of P ni =1 | G : G i | fora collection of subgroups { G , . . . , G n } satisfying T ni =1 core( G i ) = { } , wherecore( G i ) = T g ∈ G G gi .We first give a theorem due to Karpilovsky [3] which will be needed later.The proof of it can be found in [2] or [6]. Theorem 1.1.
Let A be a non-trivial finite abelian group and let A ∼ = A × . . . × A n be its direct product decomposition into non-trivial cyclic groups ofprime power order. Then µ ( A ) = a + . . . + a n , here | A i | = a i for each i . One of the themes of Johnson and Wright’s work was to establish condi-tions for when µ ( H × K ) = µ ( H ) + µ ( K ) (1)for finite groups H and K . The next result is due to Wright [8]. Theorem 1.2.
Let G and H be non-trivial nilpotent groups. Then µ ( G × H ) = µ ( G ) + µ ( H ) . Further in [8], Wright constructed a class of groups C with the propertythat for all G ∈ C , there exists a nilpotent subgroup G of G such that µ ( G ) = µ ( G ). It is a consequence of Thereom (1.2) that C is closed underdirect products and so (1) holds for any two groups H, K ∈ C . Wrightproved that C contains all nilpotent, symmetric, alternating and dihedralgroups, however the extent of it is still an open problem. In [1], Easdownand Praeger showed that (1) holds for all finite simple groups.The counter-example to (1) was provided by the referee in Wright’s pa-per [8] and involved subgroups of the standard wreath product C ≀ Sym (3),specifically the group G (5 , ,
3) which is a member of a class of unitary re-flection groups. We give a brief exposition on these groups now.Let m and n be positive integers, let C m be the cyclic group of order m and B = C m × . . . × C m be the product of n copies of C m . For each divisor p of m define the group A ( m, p, n ) by A ( m, p, n ) = { ( θ , θ , . . . , θ n ) ∈ B | ( θ θ . . . θ n ) m/p = 1 } . It follows that A ( m, p, n ) is a subgroup of index p in B and the symmetricgroup Sym ( n ) acts naturally on A ( m, p, n ) by permuting the coordinates. G ( m, p, n ) is defined to be the semidirect product of A ( m, p, n ) by Sym ( n ).It follows that G ( m, p, n ) is a normal subgroup of index p in C m ≀ Sym ( n )and thus has order m n n ! /p .It is well known that these groups can be realized as finite subgroups of GL n ( C ), specifically as n × n matrices with exactly one non-zero entry, whichis a complex m th root of unity, in each row and column such that the productof the entries is a complex ( m/p )th root of unity. Thus the groups G ( m, p, n )2re sometimes referred to as monomial reflection groups. For more detailson the groups G ( m, p, n ), see [4]. µ ( G (4 , , Recall that G (4 , ,
3) = A (4 , , ⋊ Sym (3), where A (4 , ,
3) = { ( θ , θ , θ ) ∈ C × C × C | θ θ θ = 1 } which is isomorphic to a product of two copies of the cyclic group of order4. Hence G (4 , , ∼ = ( C × C ) ⋊ Sym (3) . From now on, we will let G denote G (4 , , G = h x, y, a, b | x = y = b = a = 1 , xy = yx, x a = y, x b = y, y b = x − y − , b a = b − i . Since h x, y i ∼ = C × C is a proper subgroup of G we have by Theorem 1.1, that8 = µ ( h x, y i ) ≤ µ ( G ). Moreover since G is a proper subgroup of the wreathproduct W := C ≀ Sym (3), for which µ ( W ) = 12, we have the inequalities8 ≤ µ ( G ) ≤ . We will prove that in fact µ ( G ) = 12 by a sequence of lemmas. Lemma 2.1. h x , y i is the unique minimal normal subgroup of G .Proof. Observe by the conjugation action of a and b on x and y that M = h x , y i is indeed normal in G . Let N be a non-trivial normal subgroup of G so there exists an α = x i y j b k a l in N where i, j ∈ { , , , } , k ∈ { , , } , l ∈ { , } are not all zero. Itremains to show that M is contained in N . Case (a): k = l = 0.Subcase (i): i = j so α = x i y i .Then αα b = x i y i y i x − i y − i = y i ∈ N , so y − i α = x i ∈ N . But i = 0, so M ⊆ h x i , y i i . Hence M ⊆ N , as required.3ubcase (ii): i + j αα a = x i + j y i + j and we are back in Subcase (i).Subcase (iii): i + j ≡ αα b = x i − j y i . If 2 i − j i ≡ j mod 4. Then together with i + j ≡ i = 0. Therefore j is zero and α is trivial. This completes case (a). Case (b): k = 0 or l = 0.Subcase (i): l = 0 so k = 0Then αα − b = x i y j b k ( x − j y i − j b k ) − = x i + j y j − i . If i + j j − i i + j ≡ j − i ≡ i = j = 0 and so α = b k , whence h b i ∈ N . Hence b − b x = b − x − bx = y − x ∈ N and we are back in Case (a).Subcase (ii): l = 0 and k = 0.Then αα − a = x i y j b k a l ( x j y i b − k a l ) − = x i y j b k a l a − l b k x − j y − i = x p y q b k where p, q ∈ { , , , } and we are back in Subcase (i), replacing k by 2 k .Subcase (iii): k = 0 so l = 1Then αα − b = x i y j a ( x i y j a ) − b = x p y q b for some p, q ∈ { , , , } and again we are back in Subcase (i).This completes the proof.It is worth observing at this point that Lemma 2.1 tells us that anyminimal faithful representation of G is necessarily transitive. That is, anyminimal faithful collection of subgroups { G , . . . , G n } is just a single core-freesubroup. Lemma 2.2.
Elements of h x, y i b and h x, y i b have order . All other ele-ments of G have order dividing by . roof. It is a routine calculation to show that any element of the form α = x i y j b k for k nonzero has order three. Now suppose α = x i y j b k a l where l isnonzero. Then l = 1 and we have α = x p y q ( b k a ) = x p y q , for some p, q , which has order dividing 4. Therefore α has order dividing8. It is an immediate consequence that G does not contain any element oforder 6. Lemma 2.3. If L is a core-free subgroup of G then | G : L | ≥ .Proof. Suppose for a contradiction that core( L ) = { } and | G : L | < | G | = 96, | L | >
8. However, if | L | >
12 then | G : L | < µ ( G ) < µ ( G ) ≥
8. Therefore | L | = 12 and so by theclassification of groups of order 12, see [5], L is isomorphic to one of thefollowing groups L ∼ = C C × C A D T = h s, t | s = 1 , s = t , sts = s i Notice that the groups C , C × C , D and T each contain an elementof order 6 and so cannot be isomorphic to L by Lemma 2.2.Hence L is isomorphic to A and so we can find two non-commutingelements α = x i y j b k and β = x s y t b r of order three that generate it such that αβ has order two. Now αβ = x p y q b k + r for some p, q ∈ { , , , } and so k + r ≡ k = 1. Now αβ = x y x y α = x i y j b , we get respectively,( αβ ) α = y x y x . So in each case we get h x , y i ⊆ L , contradicting that L is core-free.Combining the above lemmas we find that any minimal faithful represen-tation of G is necessarily transitive and that any faithful transitive represen-tation has degree at least 12. Therefore we have 12 ≤ µ ( G ). But µ ( G ) ≤ Theorem 2.4.
The minimal faithful permutation degree of G (4 , , is . Let W = C ≀ Sym (3) be the wreath product of the cyclic group of order 4by the symmetric group on 3 letters. Observe at this point that since thebase group of W is C × C × C , and µ ( C × C × C ) = 12 by Theorem1.1, µ ( W ) = 12. Let γ , γ , γ be generators for the base group of W and let a = (23) , b = (123) be generators for Sym (3) acting coordinate-wise on thebase group. It follows that γ := γ γ γ commutes with a and b and thus liesin the centre of W . Let H = h γ i , so µ ( H ) = 4.Set x = γ − γ γ − and y = γ − γ − γ . Then it readily follows that x a = x b = y, y a = x, y b = x − y − , so that G = h x, y, a, b i is isomorphic to G (4 , , G ∩ H = { } .It now follows that W is an internal direct product of G and H . Thereforeby Theorem 2.4, we have12 = µ ( G × H ) < µ ( G ) + µ ( H ) = 16and so G and H form a counter-example to (1) of degree 12.6inally, we remark that using the result from [7] that µ ( G ( p, p, p )) = p for p a prime, it follows that µ ( G (3 , , C Sym (9) ( G (3 , , Sym (9) is a proper subgroup of G (3 , , G (2 , ,
3) as
Sym (4), it is immediate that µ ( G (2 , , eferences [1] D. Easdown and C.E. Praeger. On minimal faithful permutation repre-sentations of finite groups. Bull. Austral. Math. Soc. , 38:207–220, 1988.[2] D.L. Johnson. Minimal permutation representations of finite groups.
Amer. J. Math. , 93:857–866, 1971.[3] G.I. Karpilovsky. The least degree of a faithful representation of abeliangroups.
Vestnik Khar’kov Gos. Univ , 53:107–115, 1970.[4] P. Orlik and T. Hiroaki.
Arrangements and Hyperplanes . Springer-Verlag,1992.[5] J. Pedersen.
Groups of Small Order ∼ eclark/algctlg/small groups.html,2005.[6] N. Saunders. Minimal Faithful Permutation Representations of FiniteGroups . (Honours Thesis), University of Sydney, 2005.[7] N. Saunders.
The Minimal Degree for a Class of Finite Complex Reflec-tion Groups . Preprint, 2007.[8] D. Wright. Degrees of minimal embeddings of some direct products.