Normal Subgroups of Iterated Wreath Products of Symmetric Groups And

Home , Wreath product

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2 Preliminaries

Let G be a group. The intersection of all non-trivial normal subgroups Mon(G) of G is called the monolith of a group G. If Mon(G) 6=< 1 >, then the group G is called monolithic, and, in this case, Mon(G) is the least non-trivial normal subgroup of G.

Deﬁnition 1. The unique minimal normal subgroup is called the monolith.

According to [10, 15] by tableaux we call all possible inﬁnite tuple of the form:

(k−1) a = [ak]k∈N, where ak ∈ F un( M, Gk), k ∈ N. (1)

A group of isometries of the generalized Baire space E [6, 8, 9] over a family of sets

Mk, k ∈ Z, we denote by I according to [6, 9, 15]. We suppose that for every k ∈ N set

Mk is ﬁnite.

A group of isometries IZp of metric space of integer ring of p-adic numbers is isomorphic as a permutation group to the iterated wreath product of permutation groups of degree p.

Deﬁnition 2. We will say that the subset U < G is determined by its k-coordinate sets

[U]k, k ∈ N, if this subset consists of the most possible tables a ∈ IZp such that for arbitrary k ∈ N holds [a]k ∈ [U]k.

2 Deﬁnition 3. Let A be some set of the tableaux of form (1), then we will call by the k-projection of the set A the set [A]k = |a|k : a ∈ A. We call the set A as k-coordinate set if [A]k = A holds.

Deﬁnition 4. We denote by A|k a set of all functions ak such that [ε, . . . , ε, ak,ε...] ∈

[A]k.

Recall that such tableau action on element a = [ak]k∈N on element m = (mk)k∈N ∈ M is deﬁned by the rule on element:

(k−1) a ak( m) m = (mk )k∈N.

The set G of all such tableaux of the form (1) with such action which equipped by such multiplication rule forms a group. The neutral element is the table of form e = [...,ε,ε,...].

Deﬁnition 5. We will say that the subset U < G is determined by its k-coordinate sets [U]k, k ∈ N, if this subset consists of the most possible tables a ∈ I such that for arbitrary k ∈ N holds [a]k ∈ [U]k. If U is a splitting subgroup [15] of the group I, which is deﬁned in [15], then k-coordinate subset [U]k are subgroups of the group U, (k) (k) and [Uk] = U T U . Besides, family [U]k, where k ∈ N will generate a uniformly dense subgroup U ′ of a group U. The splitting normal subgroups of the group I were also studied in [9].

Let G be a group acting (from the right) by permutations on a set X and let H be an arbitrary group. Then the (permutational) wreath product H ≀ G is the semidirect product HX ⋋G, where G acts on the direct power HX by the respective permutations of the direct factors. The group Cp or (Cp, X) is equipped with a natural action by the left shift on X = {1,...,p}, p ∈ N. It is well known that the wreath product of permutation groups is associative. The multiplication rule of automorphisms g, h which are presented in the form of wreath recursion [7] g = (g(1), g(2),...,g(d))σg, h = (h(1), h(2),...,h(d))σh, is given by the formula:

g · h = (g(1)h(σg (1)), g(2)h(σg (2)),...,g(d)h(σg (d)))σgσh.

We deﬁne σ as (1, 2,...,p) where p is clear from context. The set X∗ is naturally the vertex set of a regular rooted tree, i.e. a connected graph without cycles and a designated vertex v0 called the root, in which two words are connected by an edge if and only if they are of form v and vx, where v ∈ X∗, x ∈ X.

3 n ∗ ∗ 0 The set X ⊂ X is called the n-th level of the tree X and X = {v0}. Denote the i-th j the vertex of X by vji. Note that the unique vertex vk,i corresponding to each word v in alphabet X. For every automorphism g ∈ AutX∗ and every word v ∈ X∗ we deﬁne the ∗ ∗ section (state) g(v) ∈ AutX of g at v by the rule: g(v)(x)= y for x,y ∈ X if and only ∗ k i if g(vx)= g(v)y. The subtree of X induced by the set of vertices ∪i=0X is denoted by X[k]. The restriction of the action of an automorphism g ∈ AutX∗ to the subtree X[l] is denoted by g(v)|X[l] . The restriction g(vij )|X[1] is called the vertex permutation (v.p.) of g at the vertex vij and denoted by gij. We refer to the endomorphism α|v restriction of g in a vertex v [7]. For example, if |X| = 2 then we just have to distinguish active vertices, i.e. the vertices for which α|v is non-trivial. We label every vertex of Xl, 0 ≤ l < k by sign 0 or 1 depending on the action of v.p. on it. The resulting vertex-labeled regular tree is an element of AutX[k]. Let us ﬁx some notations. The commutator of two group elements a and b is denoted by [a, b]= aba−1b−1. We denote by

ab = bab−1 the conjugation of a by b.

3 Normal subgroups of wreath product of symmetric groups

3.1 Normal subgroups in Sn ≀ Sm

We will notice at once, that further statements and expositions will be also for n > 3, and in the case when two our symmetric groups from wreath product are diﬀerent. .

Elements of our wreath product Sn ≀ Sn,n > 5, will present in the form of tableaux, where [a]1 – the active element of the table and a ∈ Sn, [a1, a2, ..., an]2 are the passive elements of the table, ai ∈ Sn, viz.:

[a1]1, [a1, a2, . . . , an]2.

The rule of multiplication of elements is described in book of Meldrum [13]. Let us give the following deﬁnitions.

Deﬁnition 6. The minimal number of transpositions in factorization of permutation corresponding to element [ai]j on transposition we will denote by rnk([ai]j). We set rnk(e) = 0.

4 For example, independent transpositions from the Moore generating system [4] for

An can be used for factorization of g in product of transpositions. Note that presentation of g ∈ An always has even number of transpositions.

Deﬁnition 7. The set of elements from Sn≀Sn,n > 5 of the tableaux form: [e]1, [a1, a2, . . . , an]2, satisfying the following condition

n N X rnk([ai]2) = 2k, k ∈ , (2) i=1 we will call set of type A and denote this set by e ≀ An. e e

In other words such subset of element g from Sn ≀ Sn,n > 5 belongs to type A if g e corresponds to automorphism of X[2] having on zero level trivial permutation and vertex permutations on second level forms an even permutation.

(0) Proposition 1. The set of elements of type A forms subgroup e ≀ An ⊳ Sn ≀ Sn. e f

Proof. Let us check e ≀ An is closed with respect to multiplication and existing of inverse f element. (1 a) Let a, b, c are elements of e ≀ A presented by tableaux: [e] , [a , a , . . . , a ] ; ’ n 1 1 2 n 2 e e e f [e]1, [b1, b2, . . . , bn]2; [e]1, [c1, c2,...,cn]2. Then (a · b) · c can be expressed by e e e [e]1, [(a1 ·b1)·c1, (a2 ·b2)·c2,..., (an ·bn)·cn]2, which is equal to, [e]1, [a1 ·(b1 ·c1), a2 ·(b2 · c2), . . . , an · (bn · cn)]2 because of associative of permutation multiplication. This tableau is equal to a · (b · c). So we have that multiplication is associative. e (1 b) Thee neutrale element [e] , [e, e, . . . , e] . ’ 1 2 (1 c) For every element of form [e] , [a , a , . . . , a ] exist inverse element a−1 ∈ e≀A , ’ 1 1 2 n 2 n −1 −1 −1 n n e −1 f of form: [e]1, [a1 , a2 , . . . , an ]2. Let us note that Pi=1 rnk([ai]2)= Pi=1 rnk([ai ]2)= N −1 2k, k ∈ , since rnk(ai)= rnk(ai ), 0 ≤ i ≤ n. Therefore e ≀ An is subgroup in Sn ≀ Sn. f Note that for product of permutations π1,π2 ∈ Sn following equation holds

rnk(π1 · π2)= rnk(π1)+ rnk(π2) − 2m,m ∈ N. (3) summand ’-2m’ arise because of possible simplifying of transposition in factorization of permutations π1,π2 ∈ Sn. Note that rank of resulting permutation is even.

For proving a normality of e ≀ An in Sn ≀ Sn we take an arbitrary element s ∈ Sn ≀ Sn. −1 f n n −1 Note that for s ∈ Sn ≀Sn equality of ranks holds Pi=1 rnk([si]2)= Pi=1 rnk([si ]2)= −1 k. Show that the product s · a · s belongs to e ≀ An. Denote representation by tableaux e f

5 −1 of s·a·s by [b1]1, [b1, b2, . . . , bn]2. Since of deﬁnition of e≀An we know that an element e n f N a ∈ e ≀ An satisfy the following property of parity i=1 rnk([ai]2) = 2k, k ∈ . Then f P etaking into account multiplication rule in a wreath product of groups and a property −1 n n (3) we have [b1]1 = [s1 · e · s1 ]1 = [e]1, and also Pi=1 rnk([bi]2)= Pi=1 rnk([si]2)+ n n −1 Pi=1 rnk([ai]2) − 2a + Pi=1 rnk([si ]2) − 2b = k + 2m − 2a + k − 2b = 2(k + m − a − b). Hence the sum is even number this satisfy the condition of belonging to e ≀ An. Thus, −1 f s · a · s ∈ e ≀ An. That accomplished the proof. e f Let us denote the element of e≀An having tableaux presentation [e]1, [(12), (12),e,e...,e]2 f by a. e Proposition 2. Normal closure NSn≀Sn (a) of a is e ≀ An. e e f Proof. To prove it, it is suﬃcient to show that N(a) is closed in e ≀ An with respect to f multiplication and conjugation. e

1) this element of N(a) is closed with respect of conjugation by elements of Sn ≀ Sn e −1 is the subgroup N(a) i.e. ∀s ∈ Sn ≀ Sn : s · a · s ∈ N(a). 2) multiplicatione of two elements ∀a, b ∈ N(a) : satisﬁese the closeness property a · b ∈ N(a). e e Denote an element of e ≀ An having form of the tableaux [e]1, [π1,π2,...,πn]2, where fn N π1,π2,...,πn ∈ Sn and Pi=0 rnk([πi]2) = 2k, k ∈ by aπ1,π2,...,πn . In particular a = e a(12),(12),e,...,e. We make the constructive prove showing that every element satisfying condition 7 belongs to normal clossuare of element a i.e. belongs to N(a). e e −1 (a) π1 = (i1i2),π2 = (j1j2),π3 = . . . = πn = e. Then aπ1,π2,...,πn = s · a · s ∈ N(a), e e where s has tableaux of form: [e]1, [((1i1)(2i2)), ((1j1)(2j2)),e,e...,e]2.

(b) π1 = (i1i2)(i3i4), π2 = . . . = πn = e. Then an element aπ1,π2,...,πn can be expressed as product of elements of type (a) where π2 = (1, 2): a(i1i2),(1,2),...,πn · a(i3i4),(1,2),...,πn ∈ N(a). Since An can be generated by pairs of transpositions then An j N(a). e e Analogous case is π2 = (j1j2)(j3j4),π1 = π3 = . . . = πn = e and the similar reasoning as above are applicable.

(c) If π1 = (i1i2) . . . (i2x−1i2x), π2 = (j1j2) . . . (j2y−1j2y),π3 = . . . = πn = e. Without losing the generality, we assume x > y. Since rnk(π1 +π2) = 2k, then (x−y) ≡ 0(mod2), since: aπ1,π2,...,πn = a(i1i2),(j1j2),e...,e·. . .·a(i2x−1i2x),(j2x−1j2x),e...,e·a(i2x+1i2x+2)(j2x+3j2x+4),e...,e·

. . . ·a(i2y−4i2y−3)(j2y−1j2y),e...,e ∈ N(a). Hence such elements are the product of the elements which are described above in thee items (a) and (b).

(d) If π1 = (i1i2) . . . (i4x−1i4x), π2 = . . . = πn = e. Then aπ1,π2,...,πn we can write: a(i1i2)(i3i4),e...,e · . . . · a(i4x−4i4x−3)(j4x−1j4x),e...,e ∈ N(a). Since such elements are product of elements from item (b). e

6 N (e) If πi 6= e, πj 6= e, rnk(πi + πj) = 2z, z ∈ ,πk = e, k 6= i, j. Then aπ1,π2,...,πn = −1 = s · aπ1,π2,e,...,e · s ∈ N(a), where s has tableau presentation: [(1i)(2j)]1, [e...,e]2. So such elements are conjugatede to the elements described in (c). N −1 (f) If πi 6= e, rnk(πi) = 2z, z ∈ ,πk = e, k 6= i. Then aπ1,π2,...,πn = s · aπ1,e,...,e · s ∈ N(a), where s has presentation: [(1i)]1, [e...,e]2. So such elements are conjugated to thee elements described in (d). n (g) If πi 6= e, i = 1, n, and the order of permutation is even i.e. Pi=1 rnk([πi]) = N 2z, z ∈ then aπ1,π2,...,πn is a product of the elements described in (e), (f). So the group closure is e ≀ An. f Remark 3. The previous statement holds for the normal closure N(b) of e b = ae,...,e,(i i ),e,...,e,(j j ),e,...,e, i1, i2, j1, j2 ∈ {1, 2,...,n}. e 1 2 1 2 Proof. To established the proof it is suﬃcient to show a ∈ N(b), since inverse inclusion e follows from previous Statement. e ae,...,e,(12),e,...,e,(12),e,...,e = ae,...,e,(i11i22),e,...,e,(j112j2),e,...,e· −1 ·b · (ae,...,e,(i 1i 2),e,...,e,(j 1j 2),e,...,e) ∈ N(b). e 1 2 1 2 e Suppose an expressed element has non trivial permutations on coordinates i and j. There- fore −1 a = s · ae,...,e,(12),e,...,e,(12),e,...,e · s ∈ N(b), e e where s has tableau form as [(1i)(2j)]1, [e...,e]2.

Let us denote by c an element of e≀An having the tableaux presentation [e]1, [(123),e,e,e...,e]2. e f Proposition 4. The normal closure N(c) of c is e ≀ An. e e f Proof. We use for the proving notiﬁcations of previous statements.

(a) If π1 ∈ An,π2 = . . . = πn = e then the element c has the form cπ1,e,...,e. e Therefore the element aπ1,π2,...,πn ∈ N(c) can be obtained by using of transformation of ﬁrst and second type from Proposition 2e (conjugations and multiplications) of an element c. e Since An ⊳ Sn and An is prime, then a normal closure its elements coincides with hole An. Hence an element aπ1,π2,...,πn ∈ N(c) where πi is 3-cycle, can be expressed from c by using transformations from prof of previouse Statements. e (b) If πi 6= e, πi ∈ An, but for another k 6= i we set πk = e, then

−1 ae,...,πi,...,e = s · aπ1,e,...,e · s ∈ N(c), e

7 where s is presented by the tableau: [(1i)]1, [e...,e]2. Therefore, ae,...,πi,...,e are conjugated with elements from item (a).

(c) If πi ∈ An, for every i = 1,n. Then cπ1,π2,...,πn can be expressed as a product of elements described in the item (b).

Thus, all elements of e ≀ An are constructed in the items (a)-(c), that is what accom- f plishes the proof.

Corollary 5. Previous Statement remains to be correct for the normal closure N(b) of e b = ae,...,e,(i i i ),e,...,e, i1, i2, i3 ∈ {1, 2,...,n}. e 1 2 3 Proof. To prove the statement it is suﬃcient to establish a ∈ N(b), because of the e belonging b ∈ N(a) follows from the a ∈ N(b). e e e Let us prove it.e So we have e −1 ae,...,e,(123),e,...,e = ae,...,e,(i 1i 2i 3),e,...,e · b · (ae,...,e,(i 1i 2i 3),e,...,e) ∈ N(b). 1 2 3 e 1 2 3 e Without losing the generality we can set non trivial permutations in obtained an element is on coordinate j. Therefore using the conjugation by the tableau [(1j)]1, [e,...,e]2 we obtain the mentioned above element a: e −1 a = s · ae,...,e,(123),e,...,e · s ∈ N(b). e e Thus, the required element a was expressed in N(b). e e ∞

Deﬁnition 8. An element of iterated wreath product ≀ Sni is called ﬁnite if, starting i=1 from some level, all vertex permutations of the level are trivial.

∗ A mapping g|v is automorphism of tree X and g|v calls restriction of g on word v ∗ ∗ or the state g in word v. This mapping g|v defined by the following rule g|v : X → X iff g(vx)= uy, where |x| = |y| and g(vx)= uy. ∗ Automorphism g calls finite if it exists n ≥ 1 such that g|v = e for all v ∈ X , |x|≥ n. ∗ The set of all finite automorphisms generate the group Autf X . m If H is proper normal subgroup of depth k and H ⊳ ≀i=1Sni , where k ∈ N, ni ≥ 5 Г´гВ m can tends to ∞, then subgroup H contains finite elements which have even permutations in all vertexes of k-th level states. Such permutation can be generated as m left based commutators of level permutation of ≀i=1Sni with depth no less than k. Now we are going to prove it. Let H be a proper normal subgroup with depth k.

Corollary 6. Subgroup H contains all ﬁnite elements with even level permutations.

8 Let v be vertex of k-th level of AutXm with non-trivial vertex permutation σ, without loosing the generality we set that σ = (1, 2, ..., l), σ ∈ Snk . In general case the permutation σ can consists of some cycles but for proof it is enough to consider one cycle. There- fore state of automorphism in v can be presented in form of the following tableau x = m−1 (b1, b2, ..., bl−1, bl)σ, bi ∈ ≀i=1 Sni , nk ≤ m, l = nk. Let y = (e, e, ..., e, e)σ1 , where σ1 = −1 −1 −1 −1 (1, 2, ..., l − 2), then z = yxy = (b2, b3, ..., bl−2, b1, bl−1, bl)σ, and zx = yxy x = −1 −1 −1 −1 −1 −1 −1 (a1, ..., al−2, e, e)σ, wher a1 = b2b1 , because of x = (bl , b1 , b2 , ..., bl−1)σ . Now we see that zx−1 is even permutation, moreover zx−1 = [y,x]. Consider the element of H stabilizing elements in vertexes 1, ..., l−2 i.e. the element of form g = (e, e, e, e, e)(l−1, l). Such g can be chosen from G\H with respect the restriction of action in v is such that g|v = (, e, e, e, e, e)(l − 1, l). Due to H is normal in G the −1 −1 −1 inclusion gyxy x g ∈ H holds. We chose x ∈ StG(v1) T StG(v2) T ... T StG(vl−2) and g from complementary vertex stabilizer intersection of y i.e. g ∈ StG(vl−1) T StG(vl). Now we conjugate yxy−1x−1 by such g = (e, e, e, e, e)(l − 1, l), which permutes only on trivial elements in tuple (a1, ..., al−2, e, e)σ ﬁxing all non trivial elements, this tuple presents the element yxy−1x−1 in the form of wreath recursion. −1 −1 −1 The obtained element gyxy x g = (a1, a2, ..., al−2, e, e)π we multiply on inverse element to commutator [y,x]: [x,y]gyxy−1x−1g−1 = (zx−1)−1gxyx−1y−1g−1 = −1 −1 −1 −1 −1 (zx ) gzx g = [x,y]g[y,x]g = (e, e, e, e, e)(τ), where τ = [σ1,σ]π, σ1 is permutation from ﬁrst level of state of portrait automorphism y, σ is root permutation of state of automorphism of x in v and π is root permutation of state of gyxy−1x−1g−1 in v. In terms of commutators (zx−1)−1gzx−1g−1 can be presented as [xz−1, g] = [[y,x], g]. Thus, we obtain the element [[y,x], g] having trivial states in vertexes of k + j levels and a permutation ρ on k-th level in v. Being a commutator ρ is even permutation from Snk therefore ρ ∈ Ank . In the similar way elements of arbitrary depth i having an even permutation on j-th level and trivial permutations on other levels can be constructed.

If we consider S3 ≀ S2 (active subgroup is the left) then we ﬁnd 7 proper normal subgroups between 9 normal subgroups: [[”1”, 1], [”A4”, 1], [”C2”, 1], [”C2 ×A4”, 1], [”C2 ×

C2”, 1], [”C2 × C2 × C2”, 1], [”C2 × S4”, 1], [”S4”, 2]].

Remark 7. The following isomorphism S3 ≀ S2 ≃ C2 × S4 is true.

Proof. The proof follows from direct computations.

For instance we analysis all normal subgroups of S3 ≀ S3. Computation in GAP lead us to following results and conclusions:

9 S3 ≀S3. The group S3 ≀S3 has 10 normal subgroups, where G0 = E and G1 = S3 ≀S3 = h(1, 2, 3), (1, 2), (4, 5, 6), (4, 5), (7, 8, 9), (7, 8), (1, 4, 7)(2, 5, 8)(3, 6, 9), (1, 4)(2, 5)(3, 6)i. The list of proper normal subgroups is the following:

G2 = h(7, 8, 9), (4, 5, 6), (1, 2, 3)i,

G3 = h(7, 8, 9), (4, 5, 6), (1, 2, 3), (2, 3)(8, 9), (2, 3)(5, 6)i,

G4 = h(7, 8, 9), (4, 5, 6), (1, 2, 3), (1, 4, 7)(2, 5, 8)(3, 6, 9), (2, 3)(5, 6)i,

G5 = h((7, 8, 9), (4, 5, 6), (1, 2, 3), (1, 4)(2, 6)(3, 5)(8, 9), (1, 4, 7)(2, 5, 8)(3, 6, 9), (2, 3)(5, 6)i,

G6 = h(7, 8, 9), (4, 5, 6), (1, 2, 3), (1, 4)(2, 5)(3, 6), (1, 4, 7)(2, 5, 8)(3, 6, 9), (2, 3)(5, 6)i,

G7 = h(2, 3)(5, 6)(8, 9), (7, 8, 9), (4, 5, 6), (1, 2, 3)i,

G8 = h(8, 9), (5, 6), (2, 3), (7, 8, 9), (4, 5, 6), (1, 2, 3)i,

G9 = h(1, 4, 7)(2, 5, 8)(3, 6, 9), (8, 9), (5, 6), (2, 3), (7, 8, 9), (4, 5, 6), (1, 2, 3)i.

Totally we have 8 proper normal subgroups in S3 ≀ S3. By analyzing their structures in terms of a semidirect product and determination them as the subgroup of wreath product with respect of their generators we obtain:

1) S3 ≀A3 ≃ (((C3 ×C3 ×C3)⋊(C2 ×C2))⋊C3)⋊C2, 3] here are 3 non equal isomorphic f 3 4 3 subgroups, also we compute ord(S3 ≀ A˜3) = ((6 ) : 2) · 6 = 3 · 2 = 108 · 6 = 648. 2) h(1, 2, 3), (1, 2), (4, 5, 6), (4, 5), (7, 8, 9), (7, 8), (1, 4, 7)(2, 5, 8)(3, 6, 9), (1, 4)(2, 5)(3, 6)i =

S3 ≀ S3. All subgroups, except one having the number 7, from this list can be easily identiﬁed.

We consider the following deﬁnition related to G7.

Deﬁnition 9. The set of elements from Sn ≀ Sn,n > 5 of the form: [e]1, [a1, a2, . . . , an]2, satisfying the following condition

n N X rnk([ai]2) = 2k, k ∈ , (4) i=1 we will call set of type A(0). e (0) Proposition 8. The set of elements of type A forms the subgroup E ≀ An ⊳ Sn ≀ Sn e f

Lemma 9. Normal closure NSn≀Sn (E ≀ An) is E ≀ An. f f Proof. According to Deﬁnition 9 the product of elements of passive subgroup of wreath product satisfy condition of parity.

It is easy to check that the normal closure of the subgroup E ≀ An in Sn ≀ Sn is itself f E ≀ An. f

10 Indeed, conjugation by arbitraty g ∈ Sn ≀ Sn an arbitrary permuted element does not change the parity of the element h ∈ E ≀ An. f Also, elements of this type A(0) described in Deﬁnition 9 with an even product of e all substitutions of base group satisfy the condition of belonging to the commutator subgroup B′ of wreath product passive group B described in the author’s article and in Meldrum’s book [11, 13].

Deﬁnition 10. A subgroup in Sn ≀ Sn, we call a subgroup of type Tn, if it contains f elements of the form:

1. ﬁrst type are elements of Tn having the form e ≀ An ⊂ Tn, f f 2. second type of Tn consists of elements with the tableau presentation: f [e]1, [π1 ...,πn]2, that ∀i = 1,...n : rnk(πi) = 2k +1, k ∈ N or equivalently πi ∈ Sn \An

(such elements be called elements of second type from subgroup Tn). f One easy can validates a correctness of this deﬁnition i.e. that this subgroup is the group as well as its normality.

We denote S3 ≀ S3 by H1 and we compute that [[H1 = ((((C3 × C3 × C3) ⋊ (C2 ×

C2)) ⋊ C3) ⋊ C2) ⋊ C2, 1]. Order of H1 is 1296.

We ﬁnd the structures of these 8 proper normal subgroups of S3 ≀ S3:

H2 = [(((C3 × C3 × C3) ⋊ (C2 × C2)) ⋊ C3) ⋊ C2, 3],

H3 = [((C3 × C3 × C3) ⋊ (C2 × C2)) ⋊ C3, 1],

H4 = [(C3 × C3 × C3) ⋊ (C2 × C2), 1],

H5 = [(C3 × C3 × C3) ⋊ C2, 1],

H6 = [C3 × C3 × C3, 1],

H7 = [S3 × S3 × S3, 1]].

Thus, S3 ≀ S3 contains 8 properly normal subgroups (one of them, numbered by 2, determined by the same products but which has non isomorphic three diﬀerent structures). Furthermore form of their generators we can deduce the splitability of this groups.

1) S3 ≀A3 ≃ (((C3 ×C3 ×C3)⋊(C2 ×C2))⋊C3)⋊C2, 3] here are 3 non equal isomorphic f 3 4 3 subgroups, also we compute ord(S3 ≀ A˜3) = ((6 ) : 2) · 6 = 3 · 2 = 108 · 6 = 648. Consider the three isomorphic subgroups having GAP structure description ["(((C3 x C3 x C3) : (C2 x C2)) : C3) : C2", 3 ] algebraic structure of them is the following:

(((C3 × C3 × C3) ⋊ (C2 × C2)) ⋊ C3) ⋊ C2,

11 these subgroups have order 648 and identifiers according to GAP system of this small group: id =703, 704 and 705 respectively. We will denote it by H703,H704 and H705 correspondently. To classify these groups we computate its commutator subgroups and ′ ′ centers. The subgroup H 703 and H 704 have order 324 and GAP idetifier is 160, this order is exactly half of 648. This means that commutator of these subgroups is the same therefore H703 and H704 are isomorphic but they are embedded in S3 ≀ S3 by in different ways as different copies. The center of this subgroups are trivial subgroup. The commutator subgroup of

S3 ≀ A˜3 is exactly A3 ≀ A˜3, so active subgroup has order in 2 times less then order of S3. Thus, we can conclude that H703 ≃ S3 ≀ A˜3 as well as H704 ≃ S3 ≀ A˜3. The center of S3 ≀ A˜3 is the trivial subgroup.

In H705 the homorphism from S3 in AutA˜3 has another structure of the kernel, ′ ′ therefor order of H705 in 3 time less then order of H703. 2) h(1, 2, 3), (1, 2), (4, 5, 6), (4, 5), (7, 8, 9), (7, 8), (1, 4, 7)(2, 5, 8)(3, 6, 9), (1, 4)(2, 5)(3, 6)i =

S3 ≀ S3.

3) T3 = (C3 × C3 × C3) ⋊ C2 taking into account that C3 ≃ A3 we see that in general case when G = Sn ≀ Sn this subgroup has structure T˜n ≃ (An × An ×···× An) ⋊ C2. | {zn } 4) ((C3 × C3 × C3) ⋊ (C2 × C2)) ⋊ C3 = A3 ≀ A˜3

Remark 10. If G = Sn ≀ Sm then the subgroup of type Tm has structure Tn,m ≃ f (Am × Am ×···× Am) ⋊ C2. | {zn }

The parity of product of permutations tuple (g1, g2, · · · , gn), where gi ∈ Am, on n (i) δ ∈ C2 depends on homomorphic image of δ ∈ C2 in Aut( Q Am ) as an element i=1 n (i) ⋊ ((g1, g2, · · · , gn), δ) of the semidirect product Q Am C2. i=1 We recall that a center of the group (A, X) ≀B is direct product of normal closure of center of diagonal of Z(Bn) i.e. (E × Z(△(Bn))), trivial an element, and intersection of (K) × E with Z(A) [11].

Note that Tm is generalization of diagonal subgroup. Taking into account for- e n mulated above properties of Sn ≀ Sm we conclude that E × ∆Z(Sm) [11] is trivial. But the following generalization of the diagonal of base of wreath product is a non-trivial normal subgroup. A passive group of N ⊳ A ≀ B consists of equivalence classes of an algebraic subgroup B, where conjugation by elements A ≀ B of acts imprimitive. These

12 generalization of diagonal subgroups of base of wreath product are Tm, Tn,m and E ≀ Am. f g In this generalization a passive subgroup of N ⊳ A ≀ B consists of equivalence classes of an algebraic subgroup B, where conjugation by elements A ≀ B of acts imprimitive.

These generalized diagonal subgroups are Tm and E ≀ Am. e g The intersection of kernel of action of active subgroup and its center is trivial because center of Sn is trivial and kernel of action (Sn,Nn), where Nn = {1, ..., n} is trivial because of action is faithful. Also in this case of the center of diagonal of base of wreath n product Sn ≀Sm i.e ∆Z(Sm) [11] is trivial too. Therefore usual diagonal subgroup of base n subgroup Sm from Sn ≀ Sm does not form normal subgroup. Thus, center of Sn ≀ Sm is trivial.

Theorem 11. Proper normal subgroups in Sn ≀Sm, n, m > 3, n, m 6= 4 are the following 2 types: 1. subgroups, acting only on second level (stabilizing the ﬁrst level), which are the following: e ≀ Am, e ≀ Am, Tm, e ≀ Sm; g f 2. subgroups, acting on both levels, which are the following: Sn ≀ Am and another g subgroup determined by the same semidirect product as H2, An ≀ Am, An ≀ Sm. g Thus, total number of proper normal subgroups in Sn ≀ Sm, n,m > 3,n,m 6= 4 is 8.

All of these groups are splittable groups. The monolith of Sn ≀ Sm is e ≀ Am if g m ≡ 0(mod2).

The subgroup Sn ≀ Am appears only in W = Sn ≀ Sm. g Proper normal subgroups in Sn ≀ Sm, n, m > 3, n, m 6= 4 are the following 2 types:

Thus, total number of proper normal subgroups in Sn ≀ Sm, n,m > 3,n,m 6= 4 is 8. ′ Note that subgroup An ≀ Am = W . Also Sn ≀ Sm, and An ≀ Sm is the new normal g f proper subgroup of Sn ≀ Sm relatively to the normal subgroups of An ≀ Sm which has 5 f proper subgroups.

Proof. First of all, note that all our normal subgroups contain subgroup e ≀ Am. If we g prove the normality of e ≀ Am then last two facts would implies that Sn ≀ Sm is monolithic g f group and its monolith is e ≀ Am. To prove the normality of e ≀ Am as well as An ≀ Sm and g g f Sn ≀ Sm it is suﬃcient to validate the embedednes (incorporation) of the normal closure f of its generators to the correspondent original subgroup e ≀ Am, An ≀ Sm and Sn ≀ Sm for g f f n,m > 3,n,m 6= 4. Generators of each of subgroups will be written in similar way with the generators of normal subgroups S3 ≀ S3. The proof of splits up of any these subgroups will follow from the form of their generators elements.

13 1. a) e≀Am: Its generating elements have the following form: ae,...,e,(i1i2i3),e,...,e, where i1, i2, i3 ∈ {1, 2,...,m}, since Am and [e]1, [(12), (12),e,e...,e]2 which denoted by a. e The normal closure of a is found in 2. Its coincide with e ≀ Am. Moreover, An can be e generated by all 3-cycles and all 3-cycles can be generated by conjugating (i1i2i3) by even substitutions, such substitutions are in Sm. Proposition 4 and Corollary 5 show us that the conjugated generator ae,...,e,(i1i2i3),e,...,e by elements of Sn ≀ Sm pertains to e ≀ Am too. Normality of the subgroup. Consider an arbitrary generator a, without losing the generality a can be in of the form: a = ae,...,e,(i1i2i3),e,...,e, i1, i2, i3 ∈ {1, 2,...,m} ∈ Am. Then for any element b ∈ Sn ≀ Sm, that has tableau presentation in the form: −1 −1 [π0]1, [π1,...,πn]2,π0 ∈ Sn,π1,...,πn ∈ Sm we have: [b · a · b ]1 = [π0 · e · π0 ] = [e]1; −1 −1 π0 −1 π0·π [b · a · b ]2 = [b(x) · a(x ) · b (x 0 )]2 = −1 π0 π0 −1 −1 = [π1 · e · π1 ,...,π(x ) · (i1i2i3) · π(x ) ,...,πn · e · πn ]2 = π0 π0 −1 = [e,...,e,π(x ) · (i1i2i3) · π(x ) ,e,...,e]2.

π0 Obviously, the mentioned above element pertain to e ≀ Am because of π(x ) · (i1i2i3) · π(xπ0 )−1 is an even substitution.

b) Generators of e≀Am are the following: ae,...,e,(i i i ),e,...,e, where i1, i2, i3 ∈ {1, 2,...,m} g 1 2 3 and a(i1i2),e,...,e,(i1i2),e,...,e, i1, i2 ∈ {1, 2,...,m}. The normal closure of these generators is found in Proposition 2. This closure NSn≀Sn (a) equals to e ≀ Am. e g d) Consider E ≀ Sm. Recall that two element generating set of Sm has form τ =

(1, 2), λ = (1, 2, ..., m). Without losing the generality we can consider generators for S3.

Taking in consideration mentioned above generating set the generators of e ≀ Sm have the form: ae,...,e,(i1i2i3),e,...,e, where i1, i2, i3 ∈ {1, 2,...,m} and ae,...,e,(i1i2),e,...,e, i1, i2, i3 ∈ {1, 2,...,m}. Let us verify the normality of this subgroup. It is obviously, because on the ﬁrst levels in result of the conjugation of generators by permutation of Sn always be identical substitution, and on the second level be any substitution from Sm which was required to prove in the theorem.

c) Tm: Generators for elements of ﬁrst type can be presented by following way: f ae,...,e,(i1i2i3),e,...,e, where i1, i2, i3 ∈ {1, 2,...,m} and generators for the second type: a(i1i2),...,(i3i4),...,(i5i6), i1, ..., i6 ∈ {1, 2,...,m}. For generators of ﬁrst type, the closedness in Tm with respect to conjugation by f elements of Sn ≀ Sm was shown in the item a). It remains to prove the same for the generators of second type. we consider the generators of the form: a(i i ),...,(i i ),...,(i i ), ∈ Tm. Then for any element 1 2 1 2 1 2 f

14 b ∈ Sn ≀ Sm, having tableau presentation [π0]1, [π1,...,πn]2,π0 ∈ Sn,

π1,...,πn ∈ Sm we have:

−1 −1 [b · a · b ]1 = [π0 · e · π0 ] = [e]1; (5) −1 −1 π0 −1 π0·π [b · a · b ]2 = [b(x) · a(x ) · b (x 0 )]2 = −1 −1 = [π1 · (i1i2) · π1 ,...,π1 · (i1i2) · πn ]2 =

π0 π0 −1 = [πk(x ) · (i1i2) · πk(x ) ]2.

π0 π0 −1 It is obvious that the obtained element [πk(x )·(i1i2)·πk(x ) ]2 is odd permutation, −1 therefore b · a · b belongs to the second type. This accomplished the prove for Tm. f 2. Here we prove that the groups mentioned in item 2 are normal with using the fact of normality in Sm of subgroups appearing on second level. Such groups are normal in

Sm.

a) An ≀ Am: Generators are the same as for e ≀ Am that are ae,...,e,(i i i ),e,...,e, and one g g 1 2 3 new generator of wreath product An ≀ Am with the following tableau presentation: g

[(i1i2i3)]1, [e,...,e]2, i1, i2, i3 ∈ {1, 2,...,n}.

Taking in account that An is generated by all 3-cycles and the fact that all 3-cycles can be generated by conjugating of (i1i2i3) by even substitutions π we use for conjugation elements of form [π]1, [e,...,e]2, π ∈ {1, 2,...,n}. As a result we see that normal closure of the new generators is An ≀ E. And normal closure of all generators is An ≀ Am. g b) Generating elements of Sn ≀ Am have the same presentation as for An ≀ Am except g g one additional generator presented by the next tableau:

[(i1i2)]1, [e,...,e]2, i1, i2, ∈ {1, 2,...,n}.

The last generator completes generation set of active group to generating set of Sn, its normal closure pertain to Sn ≀ E, the normal closure of (i1i2) is considered above in item c) for generators of Tm. Therefore this completed generating set generates Sn ≀ Am. f g c) The generating set of An ≀ Sm are the same as for described above e ≀ Sm except one additional generator in the normal closure which has tableau presentation:

[(i1i2i3)]1, [e,...,e]2, i1, i2, i3 ∈ {1, 2,...,n}.

As it is mentioned above An is generated by all 3-cycles and the fact that all 3-cycles can be generated by conjugating of (i1i2i3) by even permutations. Therefore normal closure of additional generator [(i1i2i3)]1, [e,...,e]2, i1, i2, i3 ∈ {1, 2,...,n}.

15 The example of S5 ≀ S5 conﬁrms last statement because of it give us 8 proper normal subgroups too.

Remark 12. We have to note that An ≀ Am is not normal as in Sn ≀ Sm as well as in

An ≀ Sm because of wreath product having speciﬁc conjugation as a result of action.

Theorem 13. Proper normal subgroups in An ≀Sm, n, m > 3, n, m 6= 4 are the following 3 types: 1. subgroups, acting only on second level (stabilizing the ﬁrst level), which are the following: E ≀ Am, Tm, E ≀ Sm; g f 2. subgroups, acting on both levels: An ≀ Am g 3. subgroups with active vertexes only on ﬁrst level: An ≀ E.

Thus we have 5 in An ≀ Sm and 8 proper normal subgroups in Sn ≀ Sm (active in the left).

Proof. The proofs for all subgroups except one is similar to proofs in Theorem 11. The proof of normality for An ≀ Am is such: g An ≀ Am: Its generators are the same as for e ≀ Am that are ae,...,e,(i i i ),e,...,e, and one g g 1 2 3 new generator of wreath product An ≀ Am with the following tableau presentation: g

[(i1i2i3)]1, [e,...,e]2, i1, i2, i3 ∈ {1, 2,...,n}.

Taking in account that An is generated by all 3-cycles and the fact that all 3-cycles can be generated by conjugating of (i1i2i3) by even substitutions π we use for conjugation elements of form [π]1, [e,...,e]2, π ∈ {1, 2,...,n}. As a result we see that normal closure of the last kind of generators is An≀E. And normal closure of all generators is An≀Am. g

Remark 14. Thus we have 8 proper normal subgroups in Sn ≀Sm and 5 in An ≀Sm (active in the left). The list of new proper subgroups in E ≀ Sm, Sn ≀ Am, An ≀ Sm. Thus, we have g 3 new proper normal subgroups.

In the next theorem active subgroup of wreath product is left.

Theorem 15. There are exactly 7 normal subgroups in the wreath product W = An ≀ Sn (active is left), where 5 are proper normal. The list of these subgroups is such: 1) E,

2) E ≀ An, f 3) Subgroup of type Tn that is subgroup of the ﬁrst level stabilizer StW (1) whose f elements having form g = (g1, g2, g3,...,gn) then either all gi ∈ An, or all gi ∈ Sn \ An,

16 For instance W = S5≀A5 we have a subgroup of type Tn that is subgroup of St(1) whose f elements having form g = (g1, g2, g3, g4, g5) then either all gi ∈ A5, or all gi ∈ S5 \ A5,

4) StW ′ (1) this is subgroup of ﬁrst type and has the structure E ≀ An (in other words f it is subgroup of the ﬁrst level stabilizer StW (1)), ′ ′ 5) W this is subgroup of second type and has type and W has the structure An ≀ An f in general case An ≀ Am (active is on the left) note that An ≀ An ⊳ An ≀ An, but An ≀ An is g f not normal in An ≀ Sn.

6) The level stabilizer StW (1) proper normal subgroup of W . In this case as well as in case if W = Sn ≀ Sn its normal subgroup StW (1) has structure Sn ≀ E, 7) W this is subgroup of second type. All of these groups are split groups, 5 subgroups are proper subgroup.

Proof. The proof of this theorem is written in the proof of Theorem 13 in terms of Kaloujnine tableaux and normal closure of their generators.

Hypothesis. If W = Sn ≀An, where active subgroup is left, then W contains 4 proper normal subgroups and 6 normal subgroup. The full list of the proper normal subgroups is the following:

1) E ≀ An,

2) An ≀ An The proof is obvious.

∞

3.2 Normal subgroups in ≀ Sni i=1

Consider the set Ni are the elements of a group in which the product of permutations of each element (from any level) has an even substitution. This set forms the subgroup Ni ∞ of ≀ Sni . Here the index i means the smallest level at which a non-identical substitution i=1 occurs and calls the depth of the subgroup Ni. Thus, the i-th level has a subdivision of ∞ j the tree T on which acts ≀ Sni on Y ni subtrees, at each level of which the product of i =1 i=1 substitutions is even.

Statement. The following subgroup embedding is valid:

N1 ⊃ N2 ⊃ N3 ⊃ ... Proof. Let 1 ≤ m ≤ k then at the i−th level, where m ≤ i ≤ k, in the elements of

Nk are identical substitutions e. Therefore, their product at the m-th level is an even substitution, which is a characteristic property of Nm, so Nk belongs to Nm.

17 ∞ Note that is, Nm is a group. The deﬁnition of §i=1Sni implies that the product of its elements from the k-th level forms as the product of permutations of elements the k-th level. It means that product of two element with the even product (2) is an element satisfying the property (2) too.

The presence of the inverse element in such a set Nm can be proved by contradiction. ∞

Statement. Each subgroup Ni is a normal subgroup in ≀ Sni . i=1 Obviously, because the conjugation at each level can only add an even number of odd permutations. (Moreover, the conjugation does not change the cyclic type of substitution). Each substitution of the i 1st level conjugates only one substitution of the same level, therefore the rule (3) is applicable. In general case conjugation by arbitrary element does not change parity of product of elements of Ni. ∞

Theorem 16. Let H be a normal subgroup of W = ≀ Sni . Let us assume that N1

Deﬁne a homomorphism from W (l) onto Z2, whose kernel is H(l). Another words kernel of this homomorphism is subgroup H(l) of l-th level of group H(l). If products of transpositions that belongs to H(l) is even then image of W (l) is non trivial element 1 from Z2, if products of its transpositions is odd permutation then its image is 0. The proof will be constructed by adding elements to the group N

Let us consider two arbitrarily elements a, b ∈ ≀ Sni . Let us also consider l − th level i=1 of each of these elements. Product of permutations in a−1ba of l − th level is even if and only if their product as elements of a group Snl forms even permutation of Snl . In general case, we have to notice that conjugation by an arbitrary element cannot change the product of permutations on each level each sub-tree, that corresponds to H. Indeed, ∞ by the deﬁnition in ≀ Sni each permutation i + 1-th level conjugate one permutation of i=1 ∞ i + 1-th level of initial element. Therefore, all H are normal subgroups in ≀ Sni . i=1 ∞ ∞ Z Let’s consider bijection φ : ≀ Sni → Y 2. In other words, φ gives to each element i =1 1

18 ∞ from ≀ Sni inﬁnite sequence of zeros and ones. This bijection is homomorphism of i=1 ∞ the group. If we assume that all elements of the group ≀ Sni are equivalent if and i=1 only if parity of products of elements in all levels coincides. Since, each representative ∞ Z of the class will correspond to only one element from Y 2, we get statement of the 1 theorem.

Theorem 17. Normal subgroups, that consist of ﬁnite elements, form a lattice that is isomorphic to a lattice of subgroups of inﬁnite iterate in one side Cartesian product of cyclic groups of order 2.

Proof. Proof is obvious from the previous T heorem. ∞

Theorem 18. Each Ni is a normal subgroup of ≀ Sni . i=1

∞

Proof. We need to prove that conjugation of elements from Ni with elements from ≀ Sni i=1 belongs to Ni. Consider N1. In this case everything is obvious: on each level conjugation can add only even number of odd permutations, that will not change the parity of the product of permutations on each level. In general case, we have to notice that conjugation by an arbitrary element cannot change the product of permutations on each level each ∞ sub-tree, that corresponds to Ni. Indeed, by the deﬁnition in ≀ Sni each permutation i=1 i + 1-th level conjugate one permutation of i + 1-th level of initial element. Therefore, ∞ all Ni are normal subgroups in ≀ Sni . i=1 ∞ ∞

Let ( ≀ Sni ))l−1 is a subgroup of ≀ Sni , where identical permutations are on the i=1 i=1 ﬁrst l − 1 levels. The following theorem holds:

Consider some matrix space M over Z2, each matrix has inﬁnite number of rows. We Z l−1 assign to each depth l of some element 2 the number Qi=1 ni.

Deﬁnition Two elements of matrix space M, that is characterized by a integer l and ∞ group ≀ Sni , is ω-equivalent, if one of these elements can be obtained from another by i=1 the following transformations:

1. aloud to permute ﬁrst nl−1 columns, second nl−1 columns, etc.

2. aloud to permute ﬁrst nl − 2 groups of columns, that consist of nl−1 columns, second nl−2 groups of columns , etc. 3. aloud to permute ﬁrst n groups of columns.

19 Conclusion. Further, the monolith (i.e. the unique minimal normal subgroup) of these wreath products has been investigated. The monolith exists when n ≡ 0(mod2).

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