A note on torsion length and torsion subgroups

Let 𝐾 be a group and let β : K → K be an endomorphism.

• We define the map Δ β : K → K by Δ β ⁢ ( k ) = k - 1 ⁢ β ⁢ ( k ) for all k ∈ K .

• We denote by N ⁢ ( β ) ⊲ K the normal closure of the subset

  Δ β ⁢ ( K ) = { k - 1 ⁢ β ⁢ ( k ) ∣ k ∈ K } .

• If 𝐶 is another group and ϕ : K → C is a homomorphism, we say that 𝜙 is 𝛽-neutral if ϕ ∘ β = ϕ , i.e., if ϕ ⁢ ( β ⁢ ( k ) ) = ϕ ⁢ ( k ) for all k ∈ K .

Suppose that 𝐾 is a group generated by a set 𝑋 and β ∈ End ⁢ ( K ) .

1. If ϕ : K → C is a group homomorphism, then 𝜙 is 𝛽-neutral if and only if N ⁢ ( β ) ⊆ ker ⁡ ϕ if and only if Δ β ⁢ ( X ) ⊆ ker ⁡ ϕ .

2. N ⁢ ( β ) is the normal closure in 𝐾 of the subset Δ β ⁢ ( X ) = { x - 1 ⁢ β ⁢ ( x ) ∣ x ∈ X } .

Proof

Claim (i) follows from the observation that, for each k ∈ K , ϕ ⁢ ( β ⁢ ( k ) ) = ϕ ⁢ ( k ) is equivalent to ϕ ⁢ ( k - 1 ⁢ β ⁢ ( k ) ) = 1 and the fact that two homomorphisms from 𝐺 to 𝐶 are equal if and only if they agree on 𝑋.

For claim (ii), let C = K / N , where 𝑁 is the normal closure of Δ β ⁢ ( X ) in 𝐾, and let ϕ : K → C be the natural epimorphism. Then, according to claim (i), 𝜙 is 𝛽-neutral and N ⁢ ( β ) ⊆ ker ⁡ ϕ = N . Clearly, N ⊆ N ⁢ ( β ) by definition; hence N ⁢ ( β ) = N . ∎

Let K , H be two groups, let β ∈ Aut ⁢ ( K ) be an automorphism of 𝐾, and let ϕ : K → Aut ⁢ ( H ) be a homomorphism. Denote by 𝐴 the semidirect product H ⋊ ϕ K . We define the extension of 𝛽 to 𝐴 as the map β A : A → A , given by

Using the notation of Definition 2.3, suppose that 𝜙 is 𝛽-neutral. Then the following properties hold:

1. the map β A is an automorphism of 𝐴 which has the same order as 𝛽;

2. N ⁢ ( β A ) = ( 1 , N ⁢ ( β ) ) in 𝐴;

3. the quotient A / N ⁢ ( β A ) is naturally isomorphic to the semidirect product

   H ⋊ ϕ ¯ ( K / N ⁢ ( β ) ) ,

   where ϕ ¯ : K / N ⁢ ( β ) → Aut ⁢ ( H ) is the homomorphism induced by 𝜙.

Proof

(a) Evidently, β A is a bijection with the same order as 𝛽. Consider any ( h 1 , k 1 ) , ( h 2 , k 2 ) ∈ A , where h 1 , h 2 ∈ H and k 1 , k 2 ∈ K , and observe that

On the other hand,

Since 𝜙 is 𝛽-neutral, we see that β A ⁢ ( ( h 1 , k 1 ) ⁢ ( h 2 , k 2 ) ) = β A ⁢ ( ( h 1 , k 1 ) ) ⁢ β A ⁢ ( ( h 2 , k 2 ) ) ; hence β A is an automorphism of 𝐴.

(b) By Lemma 2.2 (i), N ⁢ ( β ) ⊆ ker ⁡ ϕ , which means that the subgroup ( 1 , N ⁢ ( β ) ) centralizes the subgroup ( H , 1 ) in 𝐴. Since N ⁢ ( β ) ⊲ K and 𝐴 is generated by ( H , 1 ) ∪ ( 1 , K ) , it follows that ( 1 , N ⁢ ( β ) ) ⊲ A . Now, according to Lemma 2.2 (ii), N ⁢ ( β A ) is the normal closure in 𝐴 of the subset

Therefore, N ⁢ ( β A ) ⊆ ( 1 , N ⁢ ( β ) ) ⊆ N ⁢ ( β A ) , i.e., N ⁢ ( β A ) = ( 1 , N ⁢ ( β ) ) .

Claim (c) is a consequence of claim (b) and the fact that N ⁢ ( β ) ⊆ ker ⁡ ϕ . ∎

It is easy to check that the image Im ⁡ Δ β is always contained in the kernel ker ⁡ Σ β .

We say the automorphism β ∈ Aut ⁢ ( F ) is exact if Im ⁡ Δ β = ker ⁡ Σ β .

Let 𝐺 be a group with a torsion-free normal subgroup F ⊲ G and an element x ∈ G of finite order n ∈ N such that G = ⟨ F , x ⟩ . Let β ∈ Aut ⁢ ( F ) denote the automorphism induced by conjugation by 𝑥. The following are equivalent:

1. 𝛽 is an exact automorphism of 𝐹;

2. if an element of the form f ⁢ x for some f ∈ F has finite order in 𝐺, then it is conjugate to 𝑥 in 𝐺;

3. 𝐺 contains exactly one conjugacy class of cyclic subgroups of order 𝑛.

Proof

By the assumptions, 𝛽 must have order m ∈ N in Aut ⁢ ( F ) , where n = m ⁢ l for some l ∈ N . Therefore, for any f ∈ F , we have

Let us show that (i) implies (ii). Suppose that 𝛽 is an exact automorphism of 𝐹 and f ⁢ x ∈ G is an element of finite order for some f ∈ F . Equation (2.1) yields that Σ β ⁢ ( f ) has finite order in 𝐹, and since 𝐹 is torsion-free, we can conclude that Σ β ⁢ ( f ) = 1 . Hence f = Δ β ⁢ ( h ) for some h ∈ F by exactness. Therefore,

so (ii) holds.

To show that (ii) implies (i), suppose that f ∈ ker ⁡ Σ β in 𝐹. Then ( f ⁢ x ) n = 1 by (2.1), so in view of (ii), there must exist an element g ∈ G such that f ⁢ x = g - 1 ⁢ x ⁢ g . Now, g = x t ⁢ h for some t ∈ Z and some h ∈ F because 𝐺 is generated by 𝐹 and 𝑥 and F ⊲ G . Hence

so f = Δ β ⁢ ( h ) . Thus β ∈ Aut ⁢ ( F ) is exact, as required.

We have shown that (i) is equivalent to (ii). The equivalence between (ii) and (iii) is an easy consequence of the observation that every cyclic subgroup of order 𝑛 in 𝐺 contains precisely one element of the form f ⁢ x for f ∈ F , which will generate this subgroup. ∎

Suppose that 𝐹 is a torsion-free group with an automorphism 𝛽 of order m ∈ N . If 𝛽 is exact, then so is β p ∈ Aut ⁢ ( F ) for any integer 𝑝 coprime to 𝑚.

Proof

Let G = F ⋊ ⟨ β ⟩ be the natural semidirect product of 𝐹 with ⟨ β ⟩ ≅ C m . We identify 𝐹 with its image in 𝐺 and let x ∈ G denote the image of 𝛽. Then F ⊲ G , G = ⟨ F , x ⟩ and x ⁢ f ⁢ x - 1 = β ⁢ ( f ) for all f ∈ F .

Take any p ∈ Z coprime to 𝑚. Since 𝑥 has order 𝑚 in 𝐺, x p also has order 𝑚, so G = ⟨ F , x p ⟩ , and conjugation by x p induces the automorphism β p on 𝐹. If β ∈ Aut ⁢ ( F ) is exact, then 𝐺 has exactly one conjugacy class of cyclic subgroups of order 𝑚 by Lemma 2.7. Therefore, the same lemma shows that β p ∈ Aut ⁢ ( F ) is also exact. ∎

The results of this subsection can be expressed in terms of the first non-abelian cohomology group of the cyclic group C n with coefficients in a group 𝐹, which we compute using the presentation 2-complex associated to ⟨ β ∣ β n ⟩ rather than the general approach described in [8, Section I.5.1]. In this setting, the kernel of Σ β is the set of 1-cocycles, and the image of Δ β is the 1-coboundaries. The automorphism 𝛽 of 𝐹 is exact if and only if the pointed set H 1 ⁢ ( C n ; F ) has just one element, and Lemma 2.7 is closely related to [8, Exercise 1 in Section I.5.1].

Let K , H be groups, let β ∈ Aut ⁢ ( K ) be a finite order automorphism. Suppose that ϕ : K → Aut ⁢ ( H ) is a 𝛽-neutral homomorphism and A = H ⋊ ϕ K is the corresponding semidirect product. If 𝐻 is torsion-free and 𝛽 is an exact automorphism of 𝐾, then its extension β A , given by Definition 2.3, is an exact automorphism of 𝐴.

Proof

Suppose that 𝛽 has order m ∈ N in Aut ⁢ ( K ) . Then β A ∈ Aut ⁢ ( A ) , and it has the same order 𝑚 by Lemma 2.4 (a). Observe that, for any h ∈ H and k ∈ K , we have

where in the last equality we used the fact that 𝜙 is 𝛽-neutral.

Now, assume that Σ β A ⁢ ( ( h , k ) ) = ( 1 , 1 ) . Then Σ β ⁢ ( k ) = 1 by (2.2), so we have k = Δ β ⁢ ( u ) , for some u ∈ K , by exactness. Lemma 2.2 (i) implies that ϕ ⁢ ( k ) = Id H , so (2.2) shows that

Thus h m = 1 , whence h = 1 as 𝐻 is torsion-free. Consequently,

Therefore, ker ⁡ Σ β A = Δ β A ⁢ ( A ) , as required. ∎

The group 𝐾, its automorphism 𝛽 and the homomorphism 𝜙 enjoy the following properties.

1. ϕ ∘ β = ϕ , i.e., 𝜙 is 𝛽-neutral.

2. The quotient K / N ⁢ ( β ) is isomorphic to C 2 × C ∞ , where C 2 is generated by the image of 𝑎 and C ∞ is generated by the image of 𝑠.

3. The automorphism 𝛽 of 𝐾 is exact.

Set H 0 = { 1 } , and let β 0 ∈ Aut ⁢ ( H 0 ) be the identity automorphism; set H 1 = K , and let β 1 ∈ Aut ⁢ ( H 1 ) be the automorphism 𝛽 defined above. Now suppose that the group H n - 1 and an order 2 automorphism

have already been constructed for some n ≥ 2 . We define the group H n as the semidirect product

where ϕ n - 1 : K → Aut ⁢ ( H n - 1 ) is essentially the homomorphism 𝜙 defined above, whose image is the cyclic subgroup ⟨ β n - 1 ⟩ .

By Lemma 2.11 (i), ϕ n - 1 is 𝛽-neutral, so we can use Definition 2.3 and Lemma 2.4 to define an automorphism β n ∈ Aut ⁢ ( H n ) as the extension β H n of 𝛽 to H n . Thus

For all n ∈ N , the group H n satisfies the following properties:

1. H n is torsion-free;

2. the quotient H n / N ⁢ ( β n ) is isomorphic to

   ( H n - 1 ⋊ ⟨ β n - 1 ⟩ ) × C ∞ ,

   where ⟨ β n - 1 ⟩ ≤ Aut ⁢ ( H n - 1 ) is the cyclic subgroup of order 2 generated by β n - 1 ;

3. β n is an exact automorphism of H n .

Proof

When n = 1 , the claims follow from the definition of H 1 and Lemma 2.11. For n ≥ 2 , (i) holds because the extension of a torsion-free group by a torsion-free group is again torsion-free, (ii) is an easy consequence of Lemma 2.4 (c) and the definition of ϕ n - 1 , and claim (iii) follows from Lemma 2.10. ∎

For every integer n ≥ 1 , there exists a polycyclic group G n of torsion length 𝑛.

Proof

For each n ≥ 0 , we define the group G n + 1 as follows:

where H n and β n ∈ Aut ⁢ ( H n ) are provided by Construction 2.12. Thus H n is a subgroup of G n + 1 of index 2. Note that H n is a polycyclic group of Hirsch length 4 ⁢ n by definition; hence so is G n + 1 .

Let us show that the torsion length of G n + 1 is n + 1 by induction on 𝑛. When n = 0 , G 1 ≅ C 2 has torsion length 1; thus we can further assume that n ≥ 1 .

Recall that H n is torsion-free and β n ∈ Aut ⁢ ( H n ) is an exact automorphism by Proposition 2.13. Therefore, according to Lemma 2.7, every non-trivial finite order element of G n + 1 must be conjugate to 𝑥. Thus Tor ⁢ ( G n + 1 ) is the normal closure of 𝑥 in G n + 1 . The definition of G n + 1 , given in (2.3), implies that the quotient G n + 1 / Tor ⁢ ( G n + 1 ) is clearly isomorphic to the quotient of H n by N ⁢ ( β n ) . So Proposition 2.13.(ii) yields that

Evidently, the torsion length of G n × C ∞ is the same as the torsion length of G n , which equals 𝑛 by induction. Therefore, the torsion length of G n + 1 is n + 1 , as claimed. ∎

It is not difficult to show that the groups H n and G n , constructed above, are, in fact, virtually abelian. A more careful analysis reveals that, for every n ≥ 1 , the group G n contains a free abelian normal subgroup F n such that G n / F n is isomorphic to the direct power ( C 2 ≀ C 2 ) n - 1 . It follows that, for each n ∈ N , G n is soluble of derived length 3; hence the (restricted) direct product × n = 1 ∞ G n is a countable soluble group of derived length 3 with infinite torsion length.

It is easy to see that the torsion length of a polycyclic group 𝐺 is at most h ⁢ ( G ) + 1 , where h ⁢ ( G ) denotes the Hirsch length of 𝐺. In particular, the torsion length of any polycyclic group is finite.