$\mathbb{Q_8} = \{\pm 1, \pm i, \pm j, \pm k\}$ 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.