On the triple tensor product of prime-power groups

Theorem 1.

Let G be a finite d-generator group of prime-power order pn with Ga⁢b of order pm and exponent pe. Then |⊗3G|⩽p(n-e)⁢d2+m⁢d. Moreover, when |G′|⩽p, the bound is attained if and only if G is isomorphic to (Cpnd)d, E1 or Gp⁢(α,α,1;1,1) with α>1.

Conjecture.

Let G be a finite d-generator group of prime-power order pn with Ga⁢b of order pm and exponent pe. Then |⊗cG|⩽p(n-e)⁢dc-1+m⁢dc-2.

Conjecture.

Let G be a finite nilpotent group of class at most c. Then exp(⊗c-1G) divides exp⁡(G).

Theorem 2.

Let G be a finite nilpotent group of class k⩾3. Then exp(⊗3G) divides exp(G)[k2]-1, where [k2] is the smallest integer greater than or equal to k2. In particular, when G has nilpotency class at most 4, exp(⊗3G) divides exp⁡(G).

Proposition 2.1 ([5]).

Let G and H be groups. The following relations hold in G⊗H for all g,g′∈G and h,h′∈H:

1. (g-1⊗h)g=(g⊗h)-1=(g⊗h-1)h,

2. g′⊗(hgh-1)=(g⊗h)g′(g⊗h)-1,

3. (ghg-1)⊗h′=(g⊗h)(g⊗h)-1h′,

4. (g⊗h)(g′⊗h′)(g⊗h)-1=(g′⊗h′)[g,h],

5. [g⊗h,g′⊗h′]=(gh⁢g-1)⊗(h′g′⁢h′⁣-1),

in which [g,h] may be interpreted as gh⁢g-1∈G or hg⁢h-1∈H.

Lemma 2.2.

Suppose G is a nilpotent group of class 3 and x,y,z,t∈G. Then the actions of G and G⊗G on G⊗G reduce to

Proof.

Apply the relations of G⊗G, which are obtained by putting G=H in Proposition 2.1. ∎

Proposition 2.3.

By the above assumptions and assuming |G′|⩽p, we have

if and only if G is isomorphic with one of the following groups:

Proof.

First suppose G is abelian and

where n1⩽n2⩽⋯⩽nd and n1+n2+⋯+nd=n=m. Then

Clearly, |G⊗G|⩽p(n1+n2+⋯+nd-1)⁢d+n1+n2+⋯+nd=p(n-e)⁢d+n and equality holds if and only if n1=n2=⋯=nd, as desired.

Assume

where n1⩽n2⩽⋯⩽nd and n1+n2+⋯+nd=n-1=m. Then, from [5, Proposition 9], we have |G⊗G|⩽|Ga⁢b⊗Ga⁢b|⁢|G′⊗G|. On the other hand, as above, we have |Ga⁢b⊗Ga⁢b|⩽p(n-e-1)⁢d+n-1, and equality holds if and only if n1=n2=⋯=nd. Also, since G′⩽Z⁢(G), G′ and G act on each other trivially. So, from [6, Proposition 2.4], we have G′⊗G≅G′⊗Ga⁢b≅(Cp)d. Hence |G⊗G|⩽p(n-e)⁢d+n-1, and obviously we must have n1=n2=⋯=nd. By this hypothesis, it follows that |G⊗G|=pn⁢d. But, from [14, Theorem 2.3], we know |G⊗G|⩽p(n-1)⁢d+2, which implies that only 2-generator groups may satisfy our condition.

Thanks to [16, Theorem 36], when G is a 2-generator group of order p2⁢α+1 with Ga⁢b≅Cpα×Cpα, and p>2,

So if |G⊗G|=p2⁢n, the equality 2⁢n=4⁢α+2⁢ρ implies that ρ=1, and consequently σ=1 for γ=1. It is also readily seen that the second case does not satisfy our condition because it implies ρ>1, which contradicts the fact that ρ⩽γ=1. Thus, when p>2, only the group Gp⁢(α,α,1;1,1) has tensor square of order p2⁢n. As above, one can conclude from [16, Theorem 47] that, when p=2, only the groups G2⁢(1,1,1;0,0) and G2⁢(α,α,1;1,1) with α>1 have tensor square of order 22⁢n. From the presentation of the group Gp⁢(α,β,γ;ρ,σ), it follows that G2⁢(1,1,1;0,0)≅Q8 and Gp⁢(1,1,1;1,1)≅E1 for p>2. ∎

Lemma 2.4.

Let G be a group and x,y,z∈G, and let n be a positive integer.

1. There are homomorphisms λ:⊗3G→G⊗G and λ′:⊗3G→G given by

   λ⁢(x⊗y⊗z)=(x⊗z)z⁢(x⊗y)-1=[x,y]⊗z,

   λ′⁢(x⊗y⊗z)=[x,y,z],

   respectively, such that the actions of G on Ker⁡λ and G⊗G on Ker⁡λ′ are trivial. Furthermore, Ker⁡λ and Ker⁡λ′ are central subgroups of ⊗3G (see [6]).

2. If [x,y]⊗z=1, then x⊗y⊗zn=(x⊗y⊗z)n.

3. If [x,y,z]=1, then (x⊗y)n⊗z=(x⊗y⊗z)n.

4. If [x,y,z]∈Z⁢(G), then x⊗[y,z]n=(x⊗[y,z])n.

Proof.

The proof is straightforward. ∎

Lemma 2.5.

Let G be a nilpotent group of class 3 and x,y,z∈G. Then, for all positive integers m,n,

1. (x⁢y)n=xn⁢yn⁢[y,x](n2)⁢[y,x,x](n3)⁢[y,x,y](n2)+2⁢(n3),

2. [xm,yn]=[x,y]m⁢n⁢[x,y,x]n⁢(m2)⁢[x,y,y]m⁢(n2).

Lemma 3.1.

For any nilpotent group G of class k, we have

Proof.

Obviously, the result is true for k=2 by [1, Corollary 2.7]. Assume k=3 and a,b,x1,x2,x3∈G. The defining relations of G⊗G and Lemma 2.1 imply that

As a⁢b=[a,b]⁢b⁢a, we expand the element [x1,x2]⊗x3⊗a⁢b in two ways. First, by using Lemma 2.4 (i), we have

Note that, by Lemma 2.4 (i),

On the other hand,

Equating the terms from the two expansions yields

From here, we conclude that

So the result follows directly from equation (3.1). A similar argument to the one just presented generalizes the assertion for k>3. ∎

Proposition 3.2.

If G is a nilpotent group of class k, then the following sequence is exact:

Proof.

This follows by [9, Lemma 3] and Lemma 3.1. ∎

Lemma 3.3.

Let G be a finite d-generator nilpotent group of class at most 2. Then |⊗3G|⩽|G⊗G|d.

Proof.

Since the group G is finite, it is the direct product of its Sylow subgroups, G=S1×⋯×Sr. We can apply the distributive law in [12] to compute the tensor product ⊗3G. When i≠j, the groups Si and Sj act trivially upon each other by conjugation; therefore, for any k, the cross term Sia⁢b⊗Sja⁢b⊗Ska⁢b is trivial. Also, when i≠k, the term (Si⊗Si)a⁢b⊗Ska⁢b would be trivial. So ⊗3G≅∏i=1r⊗3Si, and it suffices to consider the case where G is a finite p-group. First suppose that G is abelian. Then G≅Cpn1×Cpn2×⋯×Cpnd, where n1⩽n2⩽⋯⩽nd. Therefore,

and

Hence

Now assume G has nilpotency class 2. Recall that G⊗G≅〈t1〉×⋯×〈tl〉 is an abelian group, where the ti’s are the generators of G⊗G (see [2]). So, for any t=t1m1⁢t2m2⁢…⁢tlml∈G⊗G and any g∈G, we have

Suppose G=〈x1,x2,…,xd〉. Then, by applying the defining relations of ⊗3G and [1], we conclude that any element tk⊗g may be presented by products of elements of the form ti⊗xj for some 1⩽i⩽l and 1⩽j⩽d. Thus

On the other hand, observe that |ti⊗xj|⩽|ti| by Lemma 2.4 (iii). Since ⊗3G is abelian, it follows that |⊗3G|⩽|t1|d|t2|d…|tr|d=|G⊗G|d, as desired. ∎

Proof of Theorem 1.

We prove the theorem by induction on n. The case n=1 is certainly true. So assume the assertion holds for all p-groups of order less than pn, and let |G|=pn. If m=0, then G is abelian, and the result holds by Lemma 3.3 and [13]. Let m≠0, |γk⁢(G)|=pr and γk+1⁢(G)=1. Since

the exact sequence given in Proposition 3.2 implies that

Now assume |G′|⩽p. By Lemma 3.3, we know that if |⊗3G|=p(n-e)⁢d2+m⁢d, then |G⊗G|=p(n-e)⁢d+m. So it suffices to verify whether the groups that appear in Proposition 2.3 attain the bound or not. Evidently, the bound is attained for G=(Cpnd)d. Assume G=Gp⁢(α,α,1;1,1)=〈a,b〉 with α⩾1. If p>2, by [16, Theorem 36], we have

Set

Thanks to [5, Proposition 10] and [1], we get ⊗3G≅(T⊗G)×(S⊗G). As T and G act on each other trivially, then T⊗G≅Ta⁢b⊗Ga⁢b≅(Cpα)6. In addition, we have (a⊗b)pα⊗a=(a⊗b⊗a)pα=1, which shows that pα divides |a⊗b⊗a|. Similarly, pα divides |a⊗b⊗b| and p divides |a⊗[a,b]⊗a|, |a⊗[a,b]⊗b|, |b⊗[a,b]⊗a| and |b⊗[a,b]⊗b|. Moreover, one can verify that the map

is a crossed pairing. Therefore, it determines a unique homomorphism

(see [5, Remark 3]), which implies that |a⊗b⊗a|=|a⊗b⊗b|=pα and |a⊗[a,b]⊗a|=|a⊗[a,b]⊗b|=|b⊗[a,b]⊗a|=|b⊗[a,b]⊗b|=p. Hence ⊗3G≅(Cpα)8×(Cp)4 which tells us that the bound is attained by G. One can analogously show that the group G2⁢(α,α,1;1,1) with α>1 attains the bound too.

Now it remains to consider the group G=Q8 with the presentation

From [16, Theorem 47], we know

Set T=〈a⊗a,b⊗b,(a⊗b)⁢(b⊗a)〉. As T⩽G⊗G, there is a homomorphism T⊗G→⊗3G which tells us |a⊗a⊗a| divides 2. On the other hand, the natural homomorphism ⊗3G→⊗3Ga⁢b implies that |a⊗a⊗a| is divisible by 2. So |a⊗a⊗a|=2. By the same manner, one can conclude that

In addition, by applying the facts

and ⊗3Ga⁢b≅(C2)8, we get |(a⊗a)⁢(b⊗b)⁢(a⊗b)⊗a|=2. Similarly, we get |(a⊗a)⁢(b⊗b)⁢(a⊗b)⊗b|=2. Thus ⊗3G≅(C2)8, whence the group Q8 does not attain the bound. ∎