The field of real numbers has the property that $\Bbb{R}^\times/\{1, -1\}$ is isomorphic to a subgroup of $\Bbb{R}^\times$, namely the positive reals. This is because there is a homomorphism $f:\ \Bbb{R}^\times \to \Bbb{R}^\times$ given by $f(x) = |x|$.
I believe there evidence to believe that $\Bbb{F}_p$ has this property as well, since $\Bbb{F}_p^\times$ is a product of cyclic groups.
Not surprisingly, I find it hard to show that the complex numbers have this property.
How do I characterize all fields with this property?
In fact $\Bbb{F}_p^{\times}$ is cyclic of order $p-1$, and hence any quotient of it is cyclic as well and its order divides $p-1$. Of course a cyclic group of order $p-1$ has a subgroup of any order dividing $p-1$, so every prime field $\Bbb{F}_p$ has this property. In fact the same argument shows that every finite field has this property.
In general, for any field $k$ the map $$k^{\times}\ \longrightarrow\ k^{\times}:\ x\ \longmapsto\ x^2,$$ is a group homomorphism with kernel $\{1,-1\}$, so the quotient $k/\{1,-1\}$ is isomorphic to its image by the first isomorphism theorem, which is a subgroup of $k^{\times}$. This is the subgroup $k^{\times2}$ of squares in $k^{\times}$. In the particular case of $k=\Bbb{C}$ we see that $\Bbb{C}/\{1,-1\}\cong\Bbb{C}$.