About ambivalent groups

In this note we shall prove some properties of the ambivalent groups and will completely determine the ambivalent solvable groups with one conjugacy class of involutions. Also we shall study the structure of the ambivalent groups having abelian Sylow 2-subgroups. All groups will be finite. The notations and definitions will be those of [3 ]and [4 ~. Definition. i) An ambivalent group is a group all whose characters are real valued. ii) A rational group is a group all of whose irreducible characters are rational valued. Proposition 1. (see [3 ]pp. 31 i) A group G is ambivalent iff for every x E G there is a t E G such that xt = x"1. ii) A group G is rational iff for every x in G the generators of x > are conjugate in G. Proposition 2.. Let G be an ambivalent group. Then : a) If N is a normal subgroup of G, then G/N is ambivalent b) Every nonidentity 2-central element of G is an involution. c~ Z(G) is an elementary abelian 2-group. d) If G is abelian, then G is an elementary abelian 2-group. e) G/G’ is an elementary abelian 2-group. f) o2~G~ = 02(~~). g) If p is an odd prime and P a Sylow p-subgroup of G, then P [P, G]. h) G is generated by its 2-elements.

i) An ambivalent group is a group all whose characters are real valued. ii) A rational group is a group all of whose irreducible characters are rational valued. Proposition 1. (see [3 ]pp. 31 i) A group G is ambivalent iff for every x E G there is a t E G such that xt = x"1. ii) A group G is rational iff for every x in G the generators of x > are conjugate in G. Proposition 2.. Let G be an ambivalent group. Then : a) If N is a normal subgroup of G, then G/N is ambivalent b) Every nonidentity 2-central element of G is an involution. c~ Z(G) is an elementary abelian 2-group. d) If G is abelian, then G is an elementary abelian 2-group. e) G/G' is an elementary abelian 2-group. f) o2~G~ = 02(~~). g) If p is an odd prime and P a Sylow p-subgroup of G, then P [P, G]. h) G is generated by its 2-elements. c) The nontrivial elements of Z(G) are 2-central. d) Follows from c).. e) Follows from d). f) Observe that )G : G' ~ = 2". g) Let FocG(P) be the focal subgroup of P in G. FocG(P) is generated by the commutators ~x, y], y E G which lie in P. Let T : G --~ P/FocG(P) stand for the transfer map. Then (see [4 Jchap. 10) T is onto and therefore P/FocG(P) is an abelian ambivalent group. Since p ~ 2, it follows that P = FocG(P). Hence P = FocG(P) P n [P, G] P and therefore P [P : G].
h) Let H be the subgroup of G generated. by the 2-elements of G. Then H is normal in G and G /H is an ambivalent group of odd order.  Let G be a solvable group and S E Sy12(G). Let x E NG(S) be an element off odd order. Then x is non-real in G.
Proof. We prove by induction on the order of G. Let N be a minimal subgroup of G. Since G is solvable, then N is an elementary abelian p-group, for a prime p. If x ~ N the image of x in G/N is non-real by induction. So x is non-real in this case. If x E N, we have that [x, S] ~ S u N = 1. Since CG(x) contains a sylow 2-subgroup of G, we have that the order of NG( x >)/CG(x) is odd. Hence x is a non-real element.
By Lemma 4 it follows immediately the following.
Let G be an ambivalent solvable group and S E Syl2(G). Then NG(S) = S.
Be G be an ambivalent solvable group and S E Syl2(G Suppose G is a solvable ambivalent group with one conjugacy class of involutions. Then, the Sylow 2-subgroups of G are isomorphic either th ?Z~ or to a generalised quaternion group. Furthermore if the Sylow 2-.subgroups of G are isomorphic to ?L2, , then G = O2, {G)~L2 inverts all the elements o f o2~ (G) and 02, , (G) = G'.
Proof. Clearly we can suppose that 02,(G) is trivial. Let I(G) = {x E Glx2 =1 ~. Let A be a minimal subgroup of G. Since G is solvable and 02(G) is trivial, it follows that A is an elementary abelian 2-group and hence A = I (G). By Thompson Theorem (see [1 ]pp. 511) if G contains more than one involution, then, Sylow 2-subgroups of G are either homocyclic or Suzuki 2-groups. Let S E Syl2(G). If S is homocyclic, then A C Z(S). If S is a Suzuki 2-group then (see [1 ]pp. 313) S' = Z(S) = A = I(S). In both cases, by Burnside Transfer Theorem, for x, y E Z(S) such that xt = y, with t, in G then there is a z E NG(S) = S such that xZ = y. Since A C Z(S) this is a contradiction. Therefore G contains only one involution, hence S is either cyclic or a generalised quaternion group. IF S is a cyclic group, by Burnside Transfer Theorem and since NG(S) = S, it follows that every element of S must be real in S, hence S is isomorphic to ZZ2.

Remark.
In [2]Feit and Seitz proved that the only groups all whose elements of same order are conjugate are the symmetric groups S1, S2 and S3. The proof of this theorem depends deeply on the classification theory of simple groups. The next corollary offer a proof of this theorem for the solvable case without using classification theory of sample groups. . Corollary 8. Let G be a solvable groups all of whose elements of the same order are conjugate. Then G is isomorphic to S1, S2 and S3.
Proof. Clearly such a group is a rational group and all involutions are conjugate in G. By theorem 7 the Sylow 2-subgroups of G are either 'lZ2 or generalized quaternion groups. Denote 02~(G) = 0(G). Case S = ?L2. By Theorem 7, G = o(G)?LZ and 0(G) = G'. If 0(G) is trivial we have the statement. Suppose 0(G) ~ 1. We shall prove now that 0(G) is an elementary abelian 3-group. Let t E S be the element of order 2. Since t inverts all elements of G' for every u, v E G' we have = uvtut-1tvt-1 = uvtuvt-1 = = 1 thus G' is abelian. Let x in G' of order Ixl = pk with p an odd prime. Since |Aut( x > ) ~ pk _ ~ ( p --1 ) and G' is abelian it follows that k = 1 and p = 3. Thus G' is an elementary abelian 3-group and since all elements of order 3 must be conjugate in G we have G' ~ ~Ls and G ~ S3.
Following [2 ]we shall prove now that the factor group of a group all whose elements of the same order are conjugate have the same property. Let N be a normal subgroup of G. Let ~, y E G/N of same order and chosse x, y E G minimal order coset representatives for ~, y. If (x~ _ ~y~ then x and y are conjugate in G and hence x and y are conjugate in G/N. Suppose |x| ~ |y| and let p a prime such that |x| = pik, ( y = pjm with i j, (p, km) = l. set u = kkm, v = ykm. Then u and v have the same order.
Clearly u and w = ' hazve both order pi hence they are conjugate in G. So u and w are conjugate in G/N hence = |w| = |v| and H = vpj-i |. It follows that p do not divide the orders of v and u. Set t = xpi and r = yp' Then t = x and j = y with contradicts the minimality of x and y. Set Z(G) = z >, z = yZ. By the previous G/ z >= H is a group with the same property as G and with abelian Sylow 2-subgroup Q8/ z >~ ?L2 x ZZ2 contradiction. Theorem 9.
Let G be an ambivalent group having abelian Sylow 2-subgroups. Let S E Sy12(G).
Then G is 2-nilpotent and splits over G' with S as complement.
Proof. By Walter [6 ], G has a normal subgroup N > 02, (G) such that G/N has off order and TV/02~(G) ~ kI x P where M is a 2-group and P is a direct product of simple groups of the form L2(q), q > 3, q = 3, 5(mod8) or q = 2n, or the Janko simple group J(11), or is of Ree type. By Prop. 2, G/N is an ambivalent group, hence G = N and M x P. Since no of the previous simple groups is ambivalent it follows that G/0~(G) ~ M. Since M is a 2-group and by Prop. 1  Conversaly, suppose that S is abelian. By theorem 2 G' = 02/ ( G), hence G' n S =1. Corollary 11. Let G be an ambivalent group with S E Syl2(G) abelian. Then S is elementary abelian and all element of G are strong real.
Proof. By theorem 9, S G/G', and the statement follows by Prop. 2. Theorem 12.
Let G be an ambivalent group having abelian Sylow 2-subgroups. Then all irreducible characters of G have Schur index 1 over R.
Proof. By Brauer-Speiser theorem (see [3 ]) 2. Let ~ E Irr(G) with = 2. From Brauer-Witt theorem (see [3 ]pp. 162) there is a subgroup W of G, W = AH (semidirect product) and a real valued irreducible character ~p of W, such that (p, is odd, where A = a > is a cyclic group of odd order and H is a 2-group. By the general properties of the Schur index mIR(03C6) = 2. If W is abelian, then cp is linear and then = 1. Thus there is some involution h E H such that ah = a-1. since W C*(a), there is a Sylow 2-subgroup T of C*(a) such that H T. Let V T, V E Syl2(C(a)) and U = AT. It is easy to see that the irreducible characters of U have Schur index 1 and degree at most 2. Let . 03BEU = 0 3 A 3 0 3 B 1 i ( 0 3 B B i + ai) + 03A3 bjvj , where a= are not real-valued and v; are real valued irreducible characters of U. All bj must be even, otherwise (çu, is odd and mIR(03BE) = l. Any irreducible character of U whose restriction to W is 03C6 is an extension of cp. Therefore , if 03BBi|U = 03C6 then 0 3 B B i | U = 0 3 C 6 hence gw = must contain c~ an even number of times which is not the case. Hence = 1.