Legendre symbol- what is the proof that it is a homomorphism? 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 $\mathbb{Z}_p$?

This is an immediately corollary of Euler's criteria:
$$(ab/p)=(ab)^{(p-1)/2}=a^{(p-1)/2}b^{(p-1)/2}$$
Euler's criteria itself is an immediate corollary of the fact that $(\mathbb{Z}/p)^\times$ is cyclic. There are many different proofs of the latter fact, but probably the most conceptually pleasing is to use structure theory of finite abelian groups.
