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 $\mathbb{Z}_p$?

I am wondering if it is permissible to use the primitive roots modulo a prime $p$.
Suppose so, and then we look at the definition of a Legendre symbol, and then give a proof that it is a homomorphism. So fix a prime $p$ first.
Let $x$ be a number not divisible by $p$. Suppose that we already have at disposal a primitive root modulo $p$, denoted by $g$. Then it is easily seen that $g$ is a non-residue of $p$, and that every integer is congruent to $g^n$ for some $n$. Moreover, $x$ is a residue if and only if $x\equiv g^{2k}$ for some $k$. Then the fact that Legendre symbol is a homomorphism follows from the rules that even+even=even, even+odd=odd, and that odd+odd=even.
Suppose we are not given the existence of a primitive root modulo $p$. Then we use another approach:
Let $p$ be an odd prime. Define a homomorphism from $\mathbb Z_p^*={1, \cdot\cdot\cdot,p-1}$ to itself by squaring: we send $x$ to $x^2$. Its kernel, that is, {$x: x^2\equiv 1\pmod p$}, consists of two numbers: $1$ and $p-1$. Its image is the set of residues modulo $p$, which is also a group $\mathbb S_p$ under multiplication. Since multiplication of integers is abelian, the group $\mathbb S_p$ is a normal subgroup of $\mathbb Z_p^*$. So we can form a quotient $\mathbb Z_p^*/\mathbb S_p$. It is of order $2$, hence isomorphic with the cyclic group of order $2$, {$1,-1$}. And we send $x\in \mathbb Z_p^*$ to its image in $\mathbb Z_p^*/\mathbb S_p$: this is the Legendre symbol. By this definition, it is obvious that Legendre symbol is a homomorphism.
I wonder also if we can talk about the wonderful interpretation due to Zolotarev: it is nothing but the theory of permutations, yet is a magic observation.  
