Question

Let a,b,,c and x elements in the group G. In each of the following solve for x in terms of a,b,c, and c.

Solve simultaneously x^2 a=bxc^-1 and acx=xac.

Answer 1

We have

$x^2a=bxc^{-1}\implies x^2ac=bxc^{-1}c=bx$

$x^2ac=x(xac)=x(acx)=bx\implies xacxx^{-1}=bxx^{-1}\implies xac=b$

$x(ac)(ac)^{-1}=b(ac)^{-1}\implies x=b(ac)^{-1}\implies x=bc^{-1}a^{-1}$