I am trying to answer the following question:
Prove that $M_{16} / Z(M_{16}) \cong C_2 \times C_2$. Where $M_{16}=\langle x, y | x^{8}=1, y^{2}=1, yx=x^{5}y\rangle$ is the modular group of order 16.
I know that $Z(M_{16})=\langle x^{2}\rangle=\{1, x^{2}, x^{4}, x^{6}\}$ and $\left( M_{16}:\langle x^{2}\rangle \right) = 4$.
I'm not sure what to do with this information to prove what I need to prove, my brain says "it's not $C_{4}$ so it must be $C_2 \times C_2$" but obviously that isn't very helpful (and probably not a correct way of thinking!).
Given what you know about the center, quotienting this group by its center is just equivalent to enforcing the additional relation $ x^2 = 1 $ in its group presentation, in which case you get an obvious presentation of the Klein four group $ C_2 \times C_2 $ - the presentation becomes
$$ \langle x, y | x^2 = 1, \, y^2 = 1, \, xy = yx \rangle $$