What are the cosets of $\mathbb{Q_8} / Z(\mathbb{Q_8})$? $\mathbb{Q_8} = \{\pm 1, \pm i, \pm j, \pm k\}$ where 1 is the identity and $\mathbb{Q_8}$ is a non-abelian group under ordinary multiplication



Let it be given that $i^2=j^2=k^2=ijk=-1$



I already calculated $Z(\mathbb{Q_8}) = \{-1, 1 \}$ by constructing a Cayley table



I want to find $\mathbb{Q_8} / Z(\mathbb{Q_8})$



Now, I know Lagrange's theorem tells us that $\mid \mathbb{Q_8} / Z(\mathbb{Q_8}) \mid = 8/2 =4$. So there must be 4 distinct cosets



Where do I go from here?
The congruence classes modulo a a normal subgroup are obtained multiplying each element of the group by all elements of the subgroup. Therefore, it is not hard to check that you obtain the four classes
$$\{1,-1\},\; \{i,-i\}, \; \{j,-j\}, \; \{k,-k\}.$$
Furthermore, the quotient $Q_8/Z$  is abelian, and each class has order $2$, so $Q_8/Z$ is isomorphic to Klein's Vierer Gruppe.
There are only two groups of order $4$.  Check the number of elements of order $2$ (or $4$) to decide which one.