Center of a Group By the center of a group $G$ we mean the set of all elements of $G$ which commute with every element of $G$, that is, $C = \{ a \in G: ax = xa \text{ for every } x \in G \}$.
We want to show that $C$ is a subgroup:
(i). Let $m, n \in C$, then $mx = xm$ and $nx = xn$ for every $x \in G$. Show that $(mn)x = x(mn)$. $mx = xm \rightarrow x = m^{-1}xm$ and $nx = xn \rightarrow x = n^{-1}xn$. Then substitute: $mnx = (mn)x = (mn)(n^{-1}xn) = mxn = m(m^{-1}xm)n = x(mn) = xmn$.
(ii). Let $m \in C$, then $mx = xm$. Show that $m^{-1}x = xm^{-1}$. From $mx = xm$ we can conclude that $x = mxm^{-1}$ and $x = m^{-1}xm$. So I need to show that $m^{-1}x = xm^{-1}$. We can substitute as follows: $m^{-1}x = m^{-1}(mxm^{-1}) = xm^{-1}$.

After solving the previous problem, I am having trouble making any progress on the following problem:
Let $C' = \{ a \in G: (ax)^{2} = (xa)^{2} \text{ for every } x \in G \}$. Prove that $C'$ is a subgroup of $G$.
(i). Let $m, n \in C'$, then $(mx)^{2} = (xm)^{2}$ and $(nx)^{2} = (xn)^{2}$. Show that $(mnx)^{2} = (xmn)^{2}$.
(ii). Let $m \in C'$, show that $m^{-1} \in C'$.
I've tried used a similar strategy to the center problem by solving for $x$ and multiplying $(mx)^{2}, (xm)^{2}, (nx)^{2}, (xn)^{2}$ in different ways. But I haven't made any progress. Could I get a hint for part (i)?
 A: For (i), write
$$
(mnx)^2 = m(nx)m(nx).
$$
Since for each $y \in G$ we have $mymy = ymym$, what can you do to the right-hand side of this? Do that, then do it again.
A: I think you would find it less confusing if you weren't using $x$ for two different things.
In (i) you know that $(mx)^2 = (xm)^2$ for all $x\in G$ and that $(nx)^2 = (xn)^2$ for all $x\in G$. You want to show that $mn\in C'$. So, let $y\in G$, and we want to show that $\Bigl((mn)y\Bigr)^2 = \Bigl(y(mn)\Bigr)^2$. Well, $(mn)y = m(ny)$; setting $x$ equal to $ny$ we have
$$\Bigl( (mn)y\Bigr)^2 = \Bigl( m(ny)\Bigr)^2 = \Bigl(mx\Bigr)^2 = \Bigl(xm\Bigr)^2 = \Bigl((ny)m\Bigr)^2.$$
Now notice that $(ny)m= n(ym)$ and do something similar. 
Likewise, for (ii), you know that $(mx)^2 = (xm)^2$ for all $x$. Take $y\in G$, and look at $(m^{-1}y)^2$. Note that
$$(m^{-1}y)^2 = \Bigl((y^{-1}m)^{-1}\Bigr)^2 = \Bigl( (y^{-1}m)^2\Bigr)^{-1}.$$
Now set $x=y^{-1}$.
P.S. And don't forget to check the sets are not empty!
A: Probably too late, for (ii)
$(gc^{-1})^2 = (egc^{-1})^2 = ((c^{-1}c)gc^{-1})^2 = (c^{-1}(cg)c^{-1})^2 = (c^{-1}(gc)c^{-1})^2 = (c^{-1}g(cc^{-1}))^2 = (c^{-1}ge)^2 = (c^{-1}g)^2$
