e-cata

Solved [math](a(bx))^{2}=((bx)a)^{2}=(b(xa))^{2}=((xa)b)^{2}=(xab)^{2}[/math] and [math](a^{-1}x)^{2}=(a^{-1}xa^{-1}a)=((a^{-1}xa^{-1})a)^{2}=(a(a^{-1}xa^{-1}))^{2}=(xa^{-1})^{2}[/math]. Thus [math]C'[/math] is closed with respect to multiplication and inverses, and is a subgroup of [math]G[/math].

This exercise is from chapter 5 of A Book Of Abstract Algebra. I've been studying this book by reading, so this is not homework. Please help. Here is my attempt: If C' is a subgroup, [math](abx)^{2}=(xab)^{2}[/math] and [math](a^{-1}x)^{2}=(xa^{-1})^{2}[/math] for every [math]a[/math] and [math]b[/math] in [math]C'[/math]. It is obvious that [math]Z(G)[/math] is included in [math]C'[/math]. If we have a group [math]K=\{a,b \in K: a=a^{-1}, ab \neq ba \}[/math], then [math]Z(K)=\emptyset[/math] even though [math](ab)^{2}=(ba)^{2}[/math] for every element in [math]K[/math]. Thus [math]C'[/math] is not necessarily Abelian.