I have some problems to understand why
$$\mathbb{Z}_{p}^{* }/(\mathbb{Z}_{p}^{* })^{2}\cong \mathbb{F}_{p}^{* }/(\mathbb{F}_{p}^{* })^{2}$$
I have seen here that there is this induced isomorphism by the fact that the canonical homomorphism $$\mathbb{Z}_{p}^{\times }\rightarrow\mathbb{F}_{p}^{\times }$$ pulls back $(\mathbb{F}_{p}^{\times })^{2}$ to $(\mathbb{Z}_{p}^{\times })^{2}$, but I can't see why this is enough. Maybe I am missing some basic property. The canonical homomorphsim is only surjective and the kernel is $1+p\mathbb{Z}_{p}$, hence I don't understand why in the quotient became an isomorphism. I think that in the quotient I have only two classes, because if $x\in\mathbb{Z}_{p}^{*}$ isn't a square, then the coset consist of all elements who's not a square. This is the same in $\mathbb{F}_{p}^{*}$ (For example I considered $\mathbb{F}_{5}^{*}$ and this works). But I'm not completely sure.