On projectively Krylov transitive and projectively weakly transitive Abelian p-groups

A group G is said to be projectively Krylov transitive if, given elements x,y∈G with UG⁢(x)=UG⁢(y), there exists ϕ∈Proj⁢(G) with ϕ⁢(x)=y.

A group G is said to be strongly projectively Krylov transitive if, given x,y∈G with UG⁢(x)=UG⁢(y), there exists ϕ∈Π⁢(G) with ϕ⁢(x)=y.

A group G is called projectively transitive if, given x,y∈G with UG⁢(x)=UG⁢(y), there exists ϕ∈Aut(G)∩Proj⁢(G) with ϕ⁢(x)=y.

A group G is called strongly projectively transitive if, given x,y∈G with UG⁢(x)=UG⁢(y), there exists ϕ∈Aut(G)∩Π⁢(G) with ϕ⁢(x)=y.

We say that the group G is projectively weakly transitive if, for any elements x,y∈G and ϕ,ψ∈Proj⁢(G) with ϕ⁢(x)=y, ψ⁢(y)=x, there exists θ∈Aut(G)∩Proj⁢(G) with θ⁢(x)=y.

We say that the group G is strongly projectively weakly transitive if, for any elements x,y∈G and ϕ,ψ∈Π⁢(G) with ϕ⁢(x)=y,ψ⁢(y)=x, there exists θ∈Aut(G)∩Π⁢(G) with θ⁢(x)=y.

If G is a p-group with p≠2, then G is projectively Krylov transitive if, and only if, G is projectively fully transitive.

Proof.

The sufficiency being elementary, we concentrate on the necessity. Reworking the proof of [19, Proposition 2.3], we observe that the same arguments work. ∎

There exists a projectively Krylov transitive 2-group G which is not projectively fully transitive and such that 2ω⁢G is a direct sum of three cyclic groups.

Proof.

Let H=〈a〉⊕〈b〉⊕〈c〉, where o⁢(a)=2, o⁢(b)=4, o⁢(c)=8. Whence |H|=64.

The lattice diagram below shows the relationships between the Ulm sequences of the elements of the group H.

Define the projections φ⁢a=a+2⁢b+4⁢c, φ⁢b=a+2⁢c, φ⁢c=a+b+7⁢c; ψ⁢a=a, ψ⁢b=0, ψ⁢c=c; α⁢a=0, α⁢b=a+b, α⁢c=0. Set Φ to be the subring of E⁢(H) generated by I,φ,ψ,α. Furthermore,

and thus all elements of the Ulm sequence (0,1,2,∞,…) are translated to elements from Φ.

Moreover,

are elements of the Ulm sequence (1,2,∞,…);

are elements of the Ulm sequence (0,1,∞,…);

are elements of the Ulm sequence (1,∞,…);

are elements of the Ulm sequence (0,∞,…);

are elements of the Ulm sequence (0,2,∞,…) and 4⁢c→𝐼4⁢c. One sees that all elements of equal Ulm sequences are sent to elements from Φ.

Taking into account the action on the element a+2⁢c, we obtain

Consequently, Φ⁢(a+2⁢c)=Ψ⁢(a+2⁢c), where Ψ is the subring of Φ, generated by I,ψ,α⁢ψ,φ⁢ψ. In view of the above equalities it is routinely observed that Ψ={x⁢I+y⁢ψ+z⁢(α⁢ψ)+t⁢(φ⁢ψ):0⩽x,y,z,t⩽8}. In fact, dropping off the coefficients in the corresponding endomorphisms, we deduce that

note that in the ⋯ space, we have products consisting of four elements, etc. Again considering the action on the element a+2⁢c, one can simply infer that

which substantiates our claim that

If now (x⁢I+y⁢ψ+z⁢(α⁢ψ)+t⁢(φ⁢ψ))⁢(a+2⁢c)=a, then we conclude that

whence x+y+t=2⁢n+1, y+z=2⁢m, x+y+3⁢t=4⁢k for some integers n,m,k which leads to the contradiction 2⁢(n+t)+1=4⁢k. Hence a∉Ψ⁢(a+2⁢c). However, (0,2,∞,…)=U⁢(a+2⁢c)<U⁢(a)=(0,∞,…), which guarantees that if G is a 2-group such that 2ω⁢G=H and E⁢(G)↾H=Φ, then G is projectively Krylov transitive but not projectively fully transitive, as wanted. ∎

It is worth noting that in the case when pω⁢G is a direct sum of two cyclic groups the situation is totally different. To demonstrate this, we first need the following technicality.

Let A=⊕i∈IAi be a group with projections πi:A→Ai, R is a ring with 1 and all πi∈R. Then:

1. if R acts Krylov transitively on Aand fully transitively on Ai for each i∈I, then R acts fully transitively on A,

2. if R acts projectively Krylov transitively on A and projectively fully transitively on Ai for each i∈I, then R acts projectively fully transitively on A.

Proof.

(1) It suffices to show the result for the direct summand ⊕j∈JAj of A, where J⊆I is a finite set, and thus by induction for the direct sum of only two groups. Let UG⁢(a1+a2)⩽UG⁢(b1+b2). According to [19, Lemma 2.2] it is enough to consider that b1+b2∈A⁢[p]. Since ht⁡(a1+a2)=min⁡{ht⁡(a1),ht⁡(a2)}, we may with no loss of generality assume that ht⁡(a1+a2)=ht⁡(a1), whence ht⁡(a1)⩽ht⁡(b1+b2)⩽ht⁡(b1). Since b1∈A1⁢[p], we have UA1⁢(a1)⩽UA1⁢(b1). Write φ⁢(a1)=b1 for some φ∈R. Observing that UA1⁢(a1)⩽UA2⁢(b2), we deduce UG⁢(a1+b2)=UG⁢(a1). So, α⁢(a1)=a1+b2 and (α-I)⁢(a1)=b2, where I is the identity of A (a unit of R). Therefore, π1⁢(a1+a2)=a1, (φ+(α-I))⁢π1∈R and (φ+(α-I))⁢π1⁢(a1+a2)=φ⁢(a1)+(α-I)⁢a1=b1+b2.

The proof of (2) is similar. ∎

Any projectively Krylov transitive group G for which pω⁢G is a direct sum of two cyclic groups is projectively fully transitive.

Proof.

The action of E⁢(G) on pω⁢G contains at least one non-trivial projection π. Then I-π is an orthogonal projection to π, so pω⁢G=π⁢(pω⁢G)⊕(I-π)⁢(pω⁢G), where each summand is cyclic. The identity map of any cyclic group acts projectively fully transitively. So, the result follows from Lemma 2.3. ∎

For every prime number p there exist strongly projectively fully transitive p-groups G with the property pω⁢G≅ℤp⊕ℤp.

Proof.

Let H=〈a〉⊕〈b〉, where o⁢(a)=o⁢(b)=p. We consider two cases.

Case (1): p⩾𝟑. Define the idempotents φ and ψ of H by φ⁢(a)=a-b, φ⁢(b)=0 and ψ⁢(a)=0, ψ⁢(b)=a+b. Let G be a group with pω⁢G≅H and E⁢(G)↾pω⁢G=Φ, where Φ is a subring of E⁢(H), generated by I,φ,ψ and I is the identity on H. What remains to show is that for all nonzero elements x⁢a+y⁢b and α⁢a+β⁢b in H there exists ϕ∈Π⁢(Φ) with the property that

If x=0 and y≠0, then

if y=0 and x≠0, then

So, let x,y≠0. Since (k⁢I+m⁢φ+n⁢ψ)⁢(x⁢a+y⁢b)=α⁢a+β⁢b, we derive

or

From the second equation we solve k=β+m⁢x-n⁢yy. Substituting k in the first equation, we obtain -(x⁢y+x2)⁢m+(y2-x⁢y)⁢n=α⁢y-β⁢x. The coefficients of m and n cannot simultaneously be zero, because if x⁢(y+x)=0 and y⁢(y-x)=0, then 2⁢x=2⁢y=0. Since p≠2, we conclude x=y=0, which contradicts our assumption. Consequently, if, for instance, x⁢y+x2≠0, then

Since n=0,…,p-1, we can find such a number m, which means that

Case (2): p=𝟐. Write H={0,a,b,a+b} and if φ⁢(a)=b, φ⁢(b)=b, ψ⁢(a)=a, ψ⁢(b)=a, θ⁢(a)=a, θ⁢(b)=0 and Φ is a subring of E⁢(H) generated by I,φ,ψ,θ. Thus we have I⁢(a)=a, φ⁢(a)=b, (I+φ)⁢(a)=a+b; ψ⁢(b)=a, I⁢(b)=b, (I+ψ)⁢b=a+b; θ⁢(a+b)=a, (I+θ)⁢(a+b)=b, I⁢(a+b)=a+b. It is now easily seen that G is a 2-group with 2ω⁢G≅H and E⁢(G)↾2ω⁢G=Φ. Hence in both cases G is a strongly projectively fully transitive group, as promised. ∎

There exists a 2-group G, which is both transitive and fully transitive but which is neither projectively transitive nor projectively fully transitive, such that 2ω⁢G is a homogeneous group of rank 4.

Proof.

Let H=〈a1〉⊕〈a2〉⊕〈a3〉⊕〈a4〉, where o⁢(aj)=2 with j=1,2,3,4. Define an endomorphism φ of the group H as follows: φ⁢a1=a2, φ⁢a2=a3, φ⁢a3=a4 and φ⁢a4=a1+a2+a3+a4. Then

Thus φ4=I+φ+φ2+φ3, where I is the identity on H. Let G be a 2-group with 2ω⁢G≅H and E⁢(G)↾2ω⁢G=Φ, where Φ is a subring of E⁢(H), generated by I,φ. Therefore, Φ={y0⁢I+y1⁢φ+y2⁢φ2+y3⁢φ3:0⩽yi⩽1,i=0,1,2,3}. It is not hard to check that if y0⁢I+y1⁢φ+y2⁢φ2+y3⁢φ3=0, then yi=0. It follows then that (x1⁢I+x2⁢φ+x3⁢φ2+x4⁢φ3)⁢a1=x1⁢a1+x2⁢a2+x3⁢a3+x4⁢a4. To prove that G is fully transitive, it suffices to show that each nonzero element x1⁢a1+x2⁢a2+x3⁢a3+x4⁢a4 will be mapped to the element a1 by the endomorphism ϕ=y0⁢I+y1⁢φ+y2⁢φ2+y3⁢φ3. In fact,

as desired. Furthermore, we obtain a system of linear equations in the variables yi, where i=0,1,2,3:

Any choice of the coefficients x1,x2,x3,x4 of this system corresponds to one of the sixteen elements x1⁢a1+x2⁢a2+x3⁢a3+x4⁢a4 of the group H.

It is also just a routine technical matter to verify that the determinant

of that system is manifestly nonzero at least for one nonzero xj. So, the given above system can be successfully solved for these xj and, by what we have just shown above, this means that G is fully transitive, indeed, as wanted.

Moreover, one sees that ker⁡ϕ=0 for every ϕ≠0, i.e., all nonzero endomorphisms of the group H are automorphisms. This also leads us to the fact that G is transitive.

Now we assume that ϕ2=ϕ. Consequently,

Hence

Taking into account that y02=y0 and y22=y2, the only nonzero solutions of that system are the following: y0=1, y1=y2=y3=0. That is why the only nonzero idempotent of the ring Φ is its identity I, so that the group G is neither projectively fully transitive nor projectively transitive, as asserted. ∎

The following statements hold.

1. Let G be a separable group such that if fn⁢(G)≠0, then fn⁢(G)⩾2 and 1∈R is a subring of E⁢(G) which acts on G Krylov transitively. Then R acts on G fully transitively.

2. Let G be a group such that pω⁢G is separable and if fn⁢(pω⁢G)≠0, then fn⁢(pω⁢G)⩾2. If G is Krylov transitive (resp., projectively Krylov transitive), then G is fully transitive (resp., projectively fully transitive).

Proof.

(1) Let y∈G⁢[p] and U⁢(x)<U⁢(y) for some x∈G (see Lemma 2.3). If ht⁡(x)<ht⁡(y), then U⁢(x+y)=U⁢(x) and thus f⁢(x)=x+y for f∈R, whence (f-I)⁢(x)=y. Assume ht⁡(x+y)>ht⁡(y)=k and hence ht⁡(x)=ht⁡(y). Let G=G0⊕G1⊕…⊕Gk⊕Gk+1⊕Gk+1*, where if Gi≠0 then Gi is a direct sum of cyclic groups of order pi for each n≥1. Since ht⁡(x)=ht⁡(y)=k, we have x,y∈Gk+1⊕Gk+1*. Let x=a+b and y=c+d, where a,c∈Gk+1 and b,d∈Gk+1*. Since ht⁡(x+y)>k, we have a=-c. So a and c are contained in the same cyclic direct summand of Gk+1. Furthermore, let z be some element from the additional direct summand of (Gk+1)⁢[p]. So, U⁢(z)=U⁢(y) and U⁢(x+z)=U⁢(x). Consequently, η⁢(x)=z for some η∈R. If now δ⁢(z)=y, then δ⁢η:x→y, as required.

Part (2) follows immediately from (1). ∎

There exists a Krylov transitive group (namely, a fully transitive 2-group) which is not projectively Krylov transitive.

Proof.

In [13, Proposition 3.5] an example was constructed of a fully transitive p-group G which is not projectively fully transitive such that (1) pω⁢G is an elementary group of infinite rank; (2) pω⁢G is an elementary p-group of rank 2 with p=5⁢n+2 or p=2; (3) p=2 and 2ω⁢G≅ℤ2⊕ℤ4. We shall construct an example of a fully transitive 2-group that is not projectively Krylov transitive for which 2ω⁢G is an elementary group of rank 3.

Let H=〈a〉⊕〈b〉⊕〈c〉, where 2⁢a=2⁢b=2⁢c=0, and define the endomorphism φ by φ⁢(a)=c, φ⁢(c)=b and φ⁢(b)=a+c. Then it is easily checked that φ3=I+φ, where I is the identity on H. Let Φ be the subring of E⁢(H) generated by I and φ. Therefore, Φ={x⁢I+y⁢φ+z⁢φ2:x,y,z=0,1}. Assuming that x⁢I+y⁢φ+z⁢φ2 is an idempotent, we get the system of comparisons mod 2, namely: x2=x, z2=y and z4+z2=z, which has only one nonzero solution x=1, y=z=0, i.e., the unique nonzero idempotent of Φ is I. Now, using Corner’s realization theorem, we exhibit a group G such that 2ω⁢G=H and E⁢(G)↾H=Φ. It is clear that E⁢(G) does not act projectively Krylov transitively on 2ω⁢G and so G is not projectively Krylov transitive by Lemma 3.1 below. However, G is fully transitive. It is enough to show that Φ acts fully transitively on H. Observe that (x⁢I+y⁢φ+z⁢φ2)⁢a=x⁢a+y⁢b+z⁢c. If z≠0, then z=1 and ((x+y)⁢I-φ)⁢(x⁢a+y⁢b+c)=(x⁢(x+y)-y)⁢a+(y⁢(x+y)-1)⁢b≠0. In fact, the system of comparisons x⁢(x+y)-y=0 and y⁢(x+y)=1 is insoluble by mod 2. Thus, it is shown that any element x⁢a+y⁢b+z⁢c≠0 is an image of a. Hereafter, (I+φ+φ2)⁢b=a and φ⁢(a+b)=a that completes the construction of the example. ∎