On necessary and sufficient conditions of simply reducibility of wreath product of finite groups

Introduction. Simply reducible groups introduced by Wigner in [1] in connection with the questions of the quantum theory have found applications in other branches of physics as well (for example, [2–6]). The questions formulated above were noted in [7–9]. A positive solution of the problem of a finite simply reducible group solvability was published in a series of papers [10–12]. Questions of the portability of various properties of groups to wreath products were considered in [13–15].

In this paper we determine necessary and sufficient conditions for the simply reducibility of the wreath product of finite groups. In the first section we find necessary and sufficient conditions for the reality of the wreath product of two finite groups. In the second section we give necessary and sufficient conditions for the simply reducibility of the wreath product of a simply reducible group and a cyclic group of order 2, and we present an infinite series of simply reducible groups built using the construction of a wreath product.

The standard group-theoretical notation is used in this paper (for example, [16]). We also use the following equivalent definition of a simply reducible group that is more convenient for computations. A finite real group is called simply reducible if

∑θ3G(g)=∑|CG(g)|2, g∈G g∈G where θG(g)=|{h∈G|h2 =g}| and CG(g) is a centralizer of the element g .

Necessary conditions for simply reducibility of the wreath product of finite groups

Theorem 1. Let H and K be finite groups, and their wreath product HsK is a simply reducible group. Then H is a real group, K is an elementary Abelian 2-group.

Proof. According to its definition, a simply reducible group is real. We show that the reality G = HsK implies that H is real, and the group K must be an elementary Abelian 2-group.

So, let the group G = HsK be real, k1=1,k2,…,kn are all the different elements of the group K. Let us take an arbitrary element h∈ Hand consider the element x = hk1 of the group G. According to the definition of the reality, there will be found such an element y=ki⋅h1k1…hnkn,h1,…,hn∈H, of the group G, that xy = x-1. We have

(h-1)k1=x-1=xy =(hn-1)kn…(h1-1)k1⋅ki-1hk1ki⋅h1k1…hnkn= =(h1-1h1)k1…(hi-1hhi)ki…(hn-1hn)kn=(hi-1hhi)ki, whence it follows that ki =1 and h-1=hi-1hhi. Thus, every element of the group H is conjugate in it with its inverse and, hence, H is a real group.

Let us prove that K is an elementary Abelian 2-group. We assume the contrary, let |kj |=s>2for some j. Without restricting the generality, we may assume that k2=kj,k3=k2j,…,ks=kjs-1.Let us consider the element g = k2 ⋅ hk1 of the group G, where 1≠h∈H.Again according to the definition of the reality there will be found such an element y=ki⋅h1k1…hnkn,h1,…,hn∈H, of the group G, that gy =g-1. We have k2-1(h-1)k1k2-1=g-1=gy=

=(hn-1)kn…(h1-1)k1ki-1k2hk1kih1k1…hnkn= kiki

=k2ki(hn-1)knk2…(h1-1)k1k2hk1kih1k1…hnkn, whence it follows that k2ki = k2-1 (in particular, ki does not lie in a cyclic subgroup 〈k2〉 ) and, therefore,

( h ^{- 1} ) ^{k 1 k 2 - 1} =( h_n ^{- 1} ) ^{knk 2 - 1} …( h ₁ ^{- 1} ) ^{k 1 k 2 - 1} h^{k 1 ki}h ₁ ^{k 1} … h_n^kn .

Remembering, that k1k2-1 =ks,      k2k2-1 =k1, k3k2-1=k2,…,ksk2-1=ks-1, we obtain (after comparing the right-hand and left-hand sides of relation (1)) the system of equations

1=h1-1hs,1=h2-1h1,1=h3-1h2,…,1=hs--11hs-2, h=hs-1hs-1, from which h=(hs-1hs-1)(hs--11hs-2)…(h3-1h2)(h2-1h1)(h1-1hs)=1, contradiction with the choice ofh. Hence, the group K is an elementary Abelian 2-group. The theorem is proved.

We show that the conditions formulated in Theorem 1 are sufficient for the realness of the group G = HsK . Let G = HsK , where the group H is real, and K is an elementary Abelian 2-group. We choose an arbitrary element g ∈ G and establish its realness.

Situation 1. Let g = h ₁ ^{k 1} … h_n^kn , where h ₁,…, h_n ∈ H . From realness H it follows that there exist such elements r ₁,…, r_n ∈ H , that h_i^ri = h_i ^{- 1} , i =1,…, n . The element y = r ₁ ^{k 1} … r_n^kn , obviously, inverts g .

Situation 2. Let g =k⋅ h1k1…hnkn , where k∈ K и h1,…, hn ∈ H. Without restricting the generality, we may assume, that the elements of the group K are arranged in such a way that kik= ki+m, i=1,…,m=n/2. According to this the element g (and any other element from K ) can be expressed by the following product m g=k⋅ ∏hikihik+ikm. i=1

We sort out for each product hihi+m, i=1,…,m, such an element fi ∈ H, that (hihi+m) fi =(hihi+m)-1 and assume m y=k⋅∏(fihi)ki(hi-1fi)kik.

i =1

Then

m gy =∏(hi-1fi-1)ki(fi-1hi)kik⋅kk× i=1

mm

× ∏ h i ^kih i ^k + ^ik m k ⋅ ∏ ( f i h i ) ^ki ( h i ^{- 1} f i ) ^kik = i =1                    i =1

m

= k ⋅ ∏ ( h i ^{- 1} f i ^{- 1} ) ^kik ( f i ^{- 1} h i ) ^kih i ^kikh i ^k + ⁱ m ( f i h i ) ^ki ( h i ^{- 1} f i ) ^kik = i =1

m

= k ⋅ ∏ ( f i ^{- 1} h i h i + m f i h i ) ^ki ( h i ^{- 1} f i ^{- 1} h i h i ^{- 1} f i ) ^kik = i =1

m

= k ⋅ ∏ ( h i ^-+ ¹ m ) ^ki ( h i ^{- 1} ) ^kik = g ^{- 1} . i =1

A sufficient condition of simply reducibility of the wreath product of a simply reducible group and a cyclic group of order 2

Theorem 2. In order that the wreath product G=HsZ 2 of a simply reducible group H and a cyclic group of order 2 be a simply reducible group it is necessary and sufficient that the equality

∑     (3 θ ⁴ H ( u )| C H ( u )| + 3 θ ² H ( u )| C H ( u )|² + | C H ( u )|³) =

( u , v ) ∈ H × H , u ∈ v^H

= 4 H ∑ C H ( h ) ² + 3 ∑ | C H ( h )| ⁴ .   (2)

h∈H                  (h,f)∈H×H,h∈fH be satisfied.

In the following three lemmas, the structure of the centralizers of the group G elements is determined.

Lemma 1. If g =(h,f), f ,h∈ H, and f v = h for some v∈ H, then

CG(g)={(t,u),σ(uv,tv-1)|t∈CH(h),u∈CH(f)}, in particular,

I C G ( g )=2 ⋅ C H ( h ) I 2_.

Proof. Let z ∈ C_G ( g ). Let us cosider two situations.

Situation 1. Let z=(x,y), x,y∈ H. Then z-1gz=(x-1,y-1)(h,f)(x,y)=(hx,fy)=(h,f), from which x ∈ CH (h) and y∈ CH (f).

Situation 2. Let z=σ(x, y),   x,y∈ H.   Then z-1=σ(y-1,x-1), z-1gz=σ(y-1,x-1)(h,f)σ(x,y)=

= (x-1,y-1)(f,h)(x,y)=(fx,hy)=(h,f), from which f x = h and hy = f. From the equalities f x = h and f v = h it follows, that x =uv, u∈ CH (f). Similarly, from the equalities hy = f and hv- = f it follows, that y = tv-1 , where t∈ CH (h). Thus, z=σ(uv,tv-1), where t∈ CH (h) and u∈ CH(f).

The second assertion of the lemma follows from the isomorphism C_H ( h ) ≅ C_H ( f ). The lemma is proved.

Lemma 2. If g =( h , f ) and the elements h and f are not conjugate in H , then

CG(g)={(x,y)|x∈CH(h),y∈CH(f)}, in particular,

I C G ( g )= C H ( h ) I ^⋅ I C H ( f ) I .

Proof. Let z∈ CG(g). If z=(x,y), then gz=(x-1,y-1)(h,f)(x,y)=(hx,fy)=(h,f).

From which x ∈ CH (h) and y∈ CH (f). If we assume that z=σ(x,y), then gz=σ(y-1,x-1)(h,f)σ(x,y)=(fx,hy)=(h,f).

which contradicts the disconjugacy of h and f . The lemma is proved.

Lemma 3. If g = σ ( h , f ) , then

C G ( g )=

={(yh,y)|y∈CH(h⋅f)}∪{σ(f-1t,th-1)|t∈CH(f⋅h)}, in particular,

I C G ( g )=2 ⋅ C H ( h ⋅ f ) I .

Proof. Let z ∈ C_G ( g ). Again we consider two situations.

Situation 1. Let z =( x , y ), x , y ∈ H . We show that z ∈ C_G ( g ) when and only when y ∈ C_H ( hf ) and x = y^h . We have

σ(h, f) =z-1gz=(x-1, y-1)σ(h, f)(x, y) = =σ(y-1,x-1)(h,f)(x,y)=σ(y-1hx,x-1fy), therefore y-1hx = h,     x-1fy=f.               (3)

Multiplying these equalities, we obtain y ^{- 1} hfy = hf , that is, y ∈ C_H ( hf ). The first equality (3) in this case gives x = y^h .

Inversely, let y e C H ( hf ) and x = y h . Then y ¹ hx = h and, consequently, x ¹ fy = h ¹ y ¹ hfy = = h ¹ hf = f , that is, equalities (3) are satisfied, hence, z - 1 gz = g .

Situation 2. Let z = c(x, y), x, y e H. Then g = z-1 gz = ^(yЛ x ^ h, f)o(x, y) =

= o ( y ⁴, x ^{- 1} )( f , h )( x , y ) = c ( y ^{- 1} fx , x ^{- 1} hy )

and the equality z ¹ gz = g is equivalent to the following two equalities:

y - fx = h ,     x ^{- 1} hy = f .             (4)

Multiplying the first equality by the second from the right, we obtain y -1 fhy = hf = ( fh) h "*, from which y = h-1t for some t e Ch (fh) and x = f -1 yh = f -1 th-1 h = f-1t.

Inversely, let x = f-1t and y = th-1 for some t e Ch (fh). Then y-1 fx = ht-1 ff-1t = h and x-1 hy = t-1 fhth-1 = fhh-1 = f, that is, equalities (4) are satisfied. The lemma is proved.

Using lemmas 1–3, we find the sum of the squares of the centralizers of the elements of the group G . We have

Z c g ( g )Г= Z C ^h , f ))I 2 + Z c g (( h , f ))I 2 = g e G              h , f e H                     h , f e H

= Z |2Ch(hf)|2 + Z    (2 ch(h)|2)2 + h, fGH                   ( h, f )e HxH, h efH

+ Z      (|C H ( h )|c h ( f )l) ² = 4| H Z c h ( h )|² +

( h , f ) e H x H , h e f^H                                      h e H

+ Z l C h ( h )| ² | C h ( f )| ² + 3     Z      c h ( h )| ⁴ . (5)

• h , f G H                              ( h , f ) e H x H , h e f^H

Lemma 4. Let g e G. Then

• 1)    6 G ( g ) = 0, if g = ⁿ ( u , v );

• 2)    6 g ( g ) = 6 h ( ^u ) 6 h ( v ), if g = ( ^u , v ) and, u and v are not conjugate in H ;

• 3)    6 _G ( g ) = 6 _H ( u ) 6 _H ( v ) + | C_H ( u )|, if u and v are conjugate in H .

Proof. The assertion 1) is obvious, since the square of any element from G lies in H x H .

Let g = ( u , v ) and z e G . If z = ( x , y ), then the equality z ² = g is satisfied when and only when u = x ²

and v = y2. If z = g(x, y), then the equality z2= g is satisfied when and only when u=yx,     v=xy.                   (6)

We show that the conditions (6) are equivalent to the fact that both u and v are conjugate in some element h e H and x = h11, y = ut 1 h, t e CH (u).             (7)

Indeed, (yx)y =xy, therefore, u and v are conjugate in H. Let v = uh. Then from (6) it follows that y = ux-1 and uh=v=xy=xux1 =ux

From which x = h ¹ 1 , y = ut ¹ h for some t e C_H ( h ).

The lemma is proved.

Using Lemma 4, we find the sum of the cubes of the values of the function 6 . We have

Z6G(g)= Z 6G (g(u,v))+ Z 6G((u,v))= geG           u, veH                  u, veH

= Z63G ((u, v)) = Z    63H (u)63H (v) + u,veH                 (u, v )e HxH, ue vH

+ Z     ( 6 h ( u ) 6 H ( v ) + l C h ( u ) l) ³ = Z 6 H ( u ) 6 H ( v ) +

( u , v ) e H x H , u e v^{H                                          u} . ^{v e H}

+ Z p

( u , v ) g H x H , u G v H ^

■ 36 H ( u )| C H ( u )| +

+ 36 H ( u )| C H ( u )|² +| C H ( u )|³

Comparing (5) with (8) and taking into account that, by virtue of the simply reducibility of the group H , the following equality is true

Z 6 H ( u ) 6 ³ h ( v ) = Z c h ( h ) 2 c h ( f )²,

• u , v e H                   h , f e H

we obtain that condition (2) is necessary and sufficient for simply reducibility of the wreath product of a simply reducible group with a cyclic group of order 2. The theorem is proved.

The proof of Theorem 2 allows us to construct the following infinite series of simply reducible 2-groups.

Corollary 1. Let G = HsZ 2 , where H is an elementary

Abelian 2-group of order 2 ⁿ . ThenG is simply reducible.

Proof. Let Z2= {1,-1} , H = Z2 x^xZ2  and h = (h1,.^,hk).   We find 6(h). Obviously, if h = (1, .,1), then 6(h) = |H| = 2”. If hi = -1 for at least one i, then 6(h) = 0, as the equation x2 = -1 is not solvable in group Z2 . As H is Abelian, then |CE (h)| = |H| for any h g H . From which,

Z

( u , v ) e H x H , u e v‘

^p 3 6 ⁴ H ( u )| C h ( u )| +

,H 1+ 3 6 H ( u )| C h ( u )| ² + | C h ( u )| ³

= 2 ⁴ n (4 + 3 ■ 2 n ) =

= 4| H I Z c h ( h )| 2 + 3     Z     c h ( h )| 4 .

h e H                ( h , f ) e H x H , h e f^H

The corollary is proved.

Conclusion. It is proved that the reality of H is the necessary condition of simply reducibility of the wreath product of the finite group H with the finite group K , and the group K must be an elementary Abelian 2-group. We also indicate sufficient conditions for simply reducibility of a wreath product of a simply reducible group with a cyclic group of order 2.

Acknowledgments. The work was supported by the Russian Foundation for Basic Research, project No. 16-01-00707a.