Subgroup of squares in $(\mathbf{Z}/p\mathbf{Z})^*$ is of index 2

Question

In a proof in my Galois Theory textbook, it proves that for $p$ an odd prime and $\eta_2$ the quadratic Gauss period in $\mathbf{Q}(\zeta_p)$, we have $f_{\mathbf{Q}}^{\eta_2}=X^2+X+\frac{1-p^*}{4}$ with $p^*=(-1)^{(p-1)/2} p$.

The proof starts with:

"Let $S\subset (\mathbf{Z}/p\mathbf{Z})^*$ be the subgroup of squares in $(\mathbf{Z}/p\mathbf{Z})^*$, and write $T=(\mathbf{Z}/p\mathbf{Z})^*\setminus S$, then we have that the quadratic Gauss period $\eta_2=\sum_{s\in S}\zeta_p^s$ ...."

It has been a long time since I took a course on group theory, so I am trying to refresh this by proving that indeed the subgroup of squares in $(\mathbf{Z}/p\mathbf{Z})^*$ is of index 2. However, I don't see why this is true.

Could someone help?

Answer 1

Since $\Bbb Z_p^\times$ is cyclic, it can be written as $$\{a^1,a^2,\ldots,a^{p-1}\}$$ The squares are exactly the elements with even exponent.

Answer 2

This results from $\mathbf Z/p\mathbf Z$ being a field, so any non-zero square modulo $p$ is the square of exactly two (opposite) elements.

Answer 3

Define a group homomorphism $$\phi:\mathbb Z_p^\times\rightarrow\mathbb Z_p^\times$$ by $\phi(x)=x^2$. We have that $\text{im}(\phi)=S$, and $\ker(\phi)=\{-1,1\}$. By the 1st isomorphism theorem $$S=\text{im}(\phi)\cong\mathbb Z_p^\times/\ker(\phi)$$ and so $|S|=\frac{p-1}{2}$. Therefore $$|\Bbb Z_p^\times:S|=2$$