Ok so there is another side to this setup.
Starting with $\mathbb{F}_p^{\times}$ take $H'$ to be the subgroup of "squares". By elementary theory this subgroup has order $\frac{p-1}{2}$. Then $H$ is the corresponding subgroup of the Galois group and $\mathbb{Q}(\sqrt{\left(\frac{-1}{p}\right)p})$ will turn out to be the corresponding fixed field.
Now $\left(\frac{q}{p}\right)=1$ is the same as saying that $q$ is a quadratic residue mod $p$, i.e. that the class of $q$ mod $p$ lies in $H'$. This means the same as $\sigma_q\in H$ by definition. But this is the same as saying $\sigma_q$ fixes the field above (by definition).
The next step of the proof is to prove that $\sigma_q$ is actually the Frobenius element of $q$ in the cyclotomic extension. The above deduction about this fixing will be equivalent to $p$ splitting completely in the quadratic field $\mathbb{Q}(\sqrt{\left(\frac{-1}{p}\right)p})$, which happens if and only if $\left(\frac{\left(\frac{-1}{p}\right)p}{q}\right) = 1$. This proves quadratic reciprocity.