On Sanov 4th-compounds of a group. (Q1409624)

Let \(M\) be a group and \(u\) be an involution in \(M\), \(S_u(M,a)\) be the group with relations \(a^2=u\) and \((ma)^4=1\) for every \(m\in M\). The group \(S_u(M,a)\) is called the Sanov compound of the group \(M\). In this article the following main results are obtained. Theorem 1. Let \(M\) be a nilpotent group. Let \(a^2=u\) be an involution in \(M\). Then \(S_u(M,a)\) is soluble. If \(M\) is finite of order \(m\), then \(S_u(M,a)\) is finite of order dividing \(2^m\cdot m\). Theorem 2. Let \(M=\langle a^2,H\rangle\), where \(H=\langle x,y:[x,y]=1\rangle\) and \(a^2\neq 1\). Let \(T_h=h^{\alpha+1}\) in \(S=S_u(M,a)\). Then: (1) \([T_x,T_y]\) is inverted by \(a\) and commutes with \(x\) and \(y\) in \(S\). (2) \([a^2,x]\) commutes with \([a^2,x]^{ay}\) in \(S\). Theorem 3. Every Sanov-compound of a non-Abelian, finite simple group is cyclic of order 2. Some examples of Sanov compounds are described.