Almost regular involutory automorphisms of uniquely 2-divisible groups

We prove that a uniquely 2-divisible group that admits an almost regular involutory automorphism is solvable.


Introduction
Recall that an automorphism ν of a group H is called involutory if ν = id and ν 2 = id. The automorphism ν is called almost regular, if C H (ν) is finite. Recall that a group U is uniquely 2-divisible if for each u ∈ U there exists a unique v ∈ U such that v 2 = u. Note that in particular a uniquely 2-divisible group contains no involutions (i.e. elements of order 2).
The purpose of this note is to use the techniques introduced in the impressive paper [Sh] of Shunkov, where he proves that a periodic group that admits an almost regular involutory automorphism is virtually solvable (i.e. it has a solvable subgroup of finite index). We prove.
Theorem 1.1. Let U be a uniquely 2-divisible group. If U admits an involutory almost regular automorphism, then U is solvable.
Our main motivation for dealing with automorphisms of uniquely 2-divisible groups comes from questions about the root groups of special Moufang sets, and those tend to be uniquely 2-divisible, see, e.g., [S]. Indeed, using Theorem 1.1 it immediately follows that Corollary 1.2. Let M(U, τ ) be a special Moufang set. If the Hua subgroup contains an involution ν such that C U (ν) is finite, then U is abelian.
We let A be a maximal abelian subgroup of U (with respect to inclusion) inverted by ν, (i.e. each element of A is inverted by ν). In Lemma 3.1(2) we show that we can take A to be infinite. We then show that for elements u 1 , . . . , u n ∈ U , the involutions u 1 νu −1 1 , . . . , u n νu −1 n in the semi-direct product U ⋊ ν invert a subgroup D ≤ A with |A : D| < ∞ (Proposition 3.5). The next step is to show that C U (D)/D is finite and solvable (Lemma 3.3).
It is easy to see that an element y ∈ U is in S iff y = νuνu −1 , for some u ∈ U , so by the above each finitely generated subgroup H of R := S is solvable and satisfies: H/Z(H) is finite. Hence R ′ is periodic (Proposition 3.6). Using the above mentioned result of Shunkov, we see that R ′ is solvable, so R is solvable.
As is well known (see [K]) U = RC U (ν) and R U . Since C U (ν) is finite and uniquely 2-divisible it has odd order. By the Feit-Thomson theorem, C U (ν) is solvable and this at last shows that U is solvable.
We remark that it is possible that with the aid of the Theorem on page 286 of [HM], one can get even more delicate information on U , but we do not need that, so we do not pursue this avenue further.

Notation and preliminary results
Notation 2.1.
(1) Throughout this note U is an infinite uniquely 2divisible group and ν ∈ Aut(U ) is an involutory automorphism which is almost regular.
(2) We denote by G the semi-direct product of U by ν and we indentify U and ν with their images in G. We let Inv(G) denote the set of involutions of G.
(4) The letter A always denotes a fixed infinite maximal (with respect to inclusion) abelian subgroup of U which is inverted by ν (i.e. all of whose elements are inverted by ν). The existence of A is guaranteed by Lemma 3.1(2) and by Zorn's lemma.
(2) Note also that for any u ∈ U , the subgroup A u is uniquely 2-divisible.
(3) It is easy to check that Lemma 2.3 ( [N], Lemma 4.1, p. 239). Let the group H be the union of finitely many, let us say n, cosets of subgroups C 1 , C 2 , . . . , C n : the index of (at least) one of these subgroups in H does not exceed n.
Corollary 2.4. Let the group H be the union of finitely many, let us say n, subsets S 1 , S 2 , . . . , S n : Then the index of (at least) one of the subgroups C 1 , . . . , C n in H does not exceed n.
Proof. For each i = 1, . . . , n, pick an arbitrary g i ∈ S i . Notice that S i ⊆ C i g i for all i, so H = n i=1 C i g i and the Corollary follows from Lemma 2.3. Lemma 2.5.
Lemma 2.6. Let D be an abelian uniquely 2-divisible subgroup of U . Then (1) C U (D)/D is a uniquely 2-divisible group.
(2) If D is inverted by ν, then νD is an almost regular involutory auto- Proof.
(1): Set C := C U (D). Assume that a, b ∈ C and a 2 D = b 2 D. Let x, y ∈ D with a 2 x = b 2 y and let u, v ∈ D with u 2 = x and v 2 = y. Then a 2 u 2 = b 2 v 2 and since a, b commute with u, v we see that (au) 2 = (bv) 2 , hence au = bv so aD = bD. Furthermore let aD ∈ C/D. Let b ∈ U with b 2 = a. Then b ∈ C and bD is the square root of aD in C/D.
(2): Clearly νD is an involutory automorphism of C/D (acting via conjugation). Assume that aD ∈ C/D centralizes νD. Then ν a = νd, for some d ∈ D. Let x ∈ D with x 2 = d. Then ν inverts x and we see that ν a = ν x and ax −1 ∈ C U (ν). It follows that C C/D (νD) = C C (ν)D/D, and since ν is almost regular, so is νD.
(3): Let xD ∈ C/D be an element inverted by νD. Then x ν = x −1 d, for some d ∈ D, and conjugating by ν we see that Now let e ∈ E. Then, by hypothesis, eD is inverted by νD, so e ν = e −1 .
3. The proof of Theorem 1.1 Lemma 3.1. Let D be an abelian subgroup of U (we allow D = 1) such that D is inverted by ν and such that C U (D) is infinite. Assume that

Then
(1) there exists an element w ∈ C U (D) D which is inverted by ν and such that C U ( D, w ) is infinite; (2) There exists an infinite abelian subgroup of U which is inverted by ν. Proof.
(1): Set V := C U (D). Then V is an infinite uniquely 2-divisible group, and ν acts on V , so without loss we may assume that U = V and that D ≤ Z(U ). Pick b ∈ S D (note that b exists by hypothesis), and write b = ντ with τ ∈ Inv(G). Let u ∈ U with u −2 = ντ, and note that since u is inverted by both ν and τ, we have ν = τ u .
We now find an element h ∈ C U (τ ) such that hu is inverted by infinitely many involutions of G. Note that hu / ∈ D; indeed, if h = 1, then hu = u and since b / ∈ D also u / ∈ D. Otherwise if hu ∈ D and h = 1, then and it follows that u inverts h which is not possible in a uniquely 2-divisible group. Since all involutions in G are conjugate, conjugating hu by an appropriate element we may assume that ν inverts hu and since hu is inverted by infinitely many involutions we see that C U (hu) is infinite and taking w = hu we are done.
It remains to show the existence of h. For each a ∈ S, let s a := ντ a and ℓ −2 a = s a . It is easy to check that since ℓ a is inverted by ν and τ a , we have τ aℓa = ν. Hence τ aℓa = τ u , and hence h a := aℓ a u −1 ∈ C U (τ ).
It follows that ℓ a = a −1 h a u. Since both ℓ a and a are inverted by ν we get after conjugating by ν that ℓ −1 a = a(h a u) ν = (h a u) −1 a. Notice now that aν ∈ Inv(G) and it follows that (h a u) aν = (h a u) −1 .
By hypothesis the set {h a | a ∈ S} is finite since it is contained in C U (τ ). Further, the set S is infinite. This implies the existence of h ∈ C U (τ ) such that the number of involutions aν that invert hu is infinite. This proves (1).
(2): If D is finite and C U (D) is infinite, then (S ∩ C U (D)) D = ∅. Hence part (2) follows from (1) by starting with D = 1 and iterating the process as long as the subgroup D, w is finite.
Lemma 3.2. Let B be a finitely generated abelian subgroup of U which is inverted by ν. Then A contains a subgroup A 1 of finite index such that A 1 , B is abelian.
Proof. Let B be a finite set of generators for B and set A 1 := b∈B A b . By Proposition 3.5 and since B is finite |A : A 1 | < ∞. Further, for each b ∈ B, ν and bνb −1 invert A 1 , so b 2 = bνb −1 ν ∈ C U (A 1 ) (recall that ν inverts b). Since U is uniquely 2-divisible, b ∈ C U (A 1 ). Hence B ≤ C U (A 1 ) and the lemma holds. Proof. Set C := C U (D) and C := C/D. Assume that C is infinite. By Lemma 2.6(1), C is uniquely 2-divisible, and by hypothesis A := A/D is a finite subgroup of C.
Let A be an infinite maximal abelian subgroup of C inverted by νD. The existence of A is guaranteed by Lemma 2.6(2) and by Lemma 3.1(2) (with C in place of U ). By Lemma 3.2 (with C in place of U and A in place of B), there exists an finite index A 1 ≤ A such that A 2 := A 1 , A is abelian. Note that A 2 is inverted by νD, so by Lemma 2.6(3), the inverse image A 2 of A 2 in C U (D) is an abelian subgroup inverted by ν. Clearly A 2 properly contains A. This contradicts the maximality of A and shows that C is finite.
Let D ≤ C be a maximal central subgroup of C which is inverted by ν. Of course D ≥ D. Further, it is clear that D is a uniquely 2-divisible group. Suppose tD is an involution in C/D. Then t 2 ∈ D, so also t ∈ D and we see that C/D has odd order. By the Feit-Thompson theorem, C/D is solvable, and the proof of the lemma is complete.
Lemma 3.4. Let x ∈ U and let s ∈ U be the unique element such that s −2 = νx −1 νx. Then xs ∈ C U (ν).
Proposition 3.5. Let A be as in Notation 2.1(4) and let u ∈ U . Let A u be as in Notation 2.1(5). Then |A : A u | < ∞.
Proof. Fix a ∈ A and consider the element This element is in U . Let s ∈ U with s −2 = νν au . By Lemma 3.4 we get that (3.1) v a := aus ∈ C U (ν).

Now set
By equation (3.1) we get s −1 = v −1 a au and conjugating by ν noticing that ν inverts a and s and centralizes v a we see that s −1 = u −ν av a . So we get the equality v −1 a au = u −ν av a , from which it follows that Let c ∈ M a , then as in equation (3.3) we get that u −1 νcv a u −1 = νv −1 a c and this together with equation (3.3) yields By equation (3.2) and by Corollary 2.4 one of the groups bc −1 | b, c ∈ M a has finite index in A, so |A : A u | < ∞ as asserted.
By the definition of A u i and by Proposition 3.5, |A : D| < ∞ and D is inverted by ν, u 1 νu −1 1 , . . . , u n νu −1 n . Also, by Remark 2.2(2), D is uniquely 2-divisible. By Lemma 3.3, C U (D)/D is finite and solvable, so since K ≤ C U (D), we see that K/Z(K) is finite and solvable. Hence (*) holds.
Next let g ∈ R ′ . Then there exists a finitely generated subgroup H of R such that g ∈ H ′ . By (*) and by [A, (33.9), p. 168], H ′ is finite, so the order of g is finite. This completes the proof of part (1).
(2): By (1), R ′ is a periodic group and since R is ν-invariant, ν is an almost regular automorphism of R ′ . By the main result of Shunkov in [Sh], R ′ is virtually solvable. But by (*), R ′ is also locally solvable, so this shows that R ′ is solvable and hence so is R.