Solved
(a(bx))^{2}=((bx)a)^{2}=(b(xa))^{2}=((xa)b)^{2}=(xab)^{2}
and
(a^{-1}x)^{2}=(a^{-1}xa^{-1}a)=((a^{-1}xa^{-1})a)^{2}=(a(a^{-1}xa^{-1}))^{2}=(xa^{-1})^{2}.
Thus C' is closed with respect to multiplication and inverses, and is a subgroup of G.
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,
(abx)^{2}=(xab)^{2}
and
(a^{-1}x)^{2}=(xa^{-1})^{2}
for every a and b in C'.
It is obvious that Z(G) is included in C'. If we have a group K=\{a,b \in K: a=a^{-1}, ab \neq ba \},
then Z(K)=\emptyset even though (ab)^{2}=(ba)^{2} for every element in K. Thus C' is not necessarily Abelian.
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