TITLE: Legendre symbol- what is the proof that it is a homomorphism?
QUESTION [6 upvotes]: I know that one property of the Legendre symbol is that it is a homomorphism. However, I have not been able to find a proof that this is the case. If someone could give me or show me to a thorough proof of this, that would be great.
I am going with the definition:
$\sigma(x) = 1$ when $ x=y^2$
$\sigma(x)= -1 $ otherwise
where $\sigma$ is a map st $\sigma: {\mathbb{Z}_p}^{\times} \rightarrow (-1,1)$

EDIT: How would we sketch a proof that the symbol is a homomorphism using that 
if $(G, *)$ is a finite group and $H\subset G$ is a subgroup, and we have an equivalence relation on $G: x \sim y$ iff $\exists h \in H$ st $y= x*h$, and $P$ is an equivalence relation: then we know $\# P = \# H$ and $\# H$ divides $\# G$. 
Basically how would we prove, using this fact, that the set of squares in $\mathbb{Z}_p$ is a subgroup of $\mathbb{Z}_p$?

REPLY [4 votes]: I feel I should add here a standard result of group theory: a subset of a finite group closed under multiplication is a subgroup. Since $(\Bbb Z_p)^{\times}$ is clearly finite, it suffices to show that a QR times a QR is a QR, and the other cases are unnecessary.
Note that the map $x \to x^2$ is an ONTO multiplicative homomorphism from $(\Bbb Z_p)^{\times}$ to the set of quadratic residues (mod $p$) which has kernel $\{1,p-1\}$. This shows that the index of the subgroup of quadratic residues is $2$, which means there must be $\dfrac{p-1}{2}$ quadratic residues (assuming $p$ is an odd prime, this formula breaks down for $p = 2$).
Since NR is thus the "other coset", the other rules for the products:
NR*NR, NR*QR
follow from the rules of coset multiplication in $(\Bbb Z_p)^{\times}/QR$