Abstract Algebra

(A)
x has order 4, so $1,x,x^2,x^3\;(=x^{-1})$ are distinct elements; they are also distinct from y (otherwise Q would be cyclic). So we have five distinct elements. And since $y^2=x^2$ , it follows that Q has at most eight elements: $1,x,x^2,x^3,y,xy,x^2y,x^3y$ . All you need to do is to show that the last three of these elements are distinct from the first five and from each other. For example, for xy:

If $xy=1$ , then $y=x^{-1}=x^3$ (contradiction). Hence $xy\ne1$ .
If $xy=x$ , then $y=1$ (contradiction). Hence $xy\ne x$ .
If $xy=x^2$ , then $y=x$ (contradiction). Hence $xy\ne x^2$ .
If $xy=x^3$ , then $y=x^2$ (contradiction). Hence $xy\ne x^3$ .
If $xy=y$ , then $x=1$ (contradiction). Hence $xy\ne y$ .

Show similarly that $x^2y,x^3y\not\in\{1,x,x^2,x^3,y\}$ , then show that $xy\ne x^2y,x^2y\ne x^3y,x^3y\ne xy$ .

(B)
We have $y\ne1,\ y^2=x^2\ne1,\ y^3=(y^2)y=x^2y\ne1$ and $y^4=(y^2)^2=(x^2)^2=x^4=1$ .

$xyx^{-1}=xy(yxy^{-1})$ … and the rest is straightforward.

(C)
For any $k\in\mathbb{Z}$ , $x^k=x^{k'}$ where $k'\in\{0,1,2,3\}$ and $k\equiv k'\pmod{4}$ .

(D)
$x^2$ is the only element of order 2. This is just a matter of checking one by one.

(E)
We have $yx=yxy^{-1}y=x^{-1}y\ne xy$ ; hence x and y do not belong to the centre, and neither do their inverses $x^3$ and $x^2y$ . Also $(xy)y=xy^2=x^3$ but $yxy=yx(y^{-1}y)y=(yxy^{-1})y^2=x^{-1}x^2=x\ne (xy)y$ . Hence, neither xy nor $(xy)^{-1}=x^3y$ belongs to the centre. That leaves $x^2$ . Simple checking should verify that $x^2$ does commute with every other element in Q. Hence $\mathrm{Z}(Q)=\{1,x^2\}$ .