(A)

xhas order 4, so are distinct elements; they are also distinct fromy(otherwiseQwould be cyclic). So we have five distinct elements. And since , it follows thatQhas at most eight elements: . 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, forxy:

- If , then (contradiction). Hence .
- If , then (contradiction). Hence .
- If , then (contradiction). Hence .
- If , then (contradiction). Hence .
- If , then (contradiction). Hence .

Show similarly that , then show that .

(B)

We have and .

… and the rest is straightforward.

(C)

For any , where and .

(D)

is the only element of order 2. This is just a matter of checking one by one.

(E)

We have ; hencexandydo not belong to the centre, and neither do their inverses and . Also but . Hence, neitherxynor belongs to the centre. That leaves . Simple checking should verify that does commute with every other element inQ. Hence .