Subgroup structure of multiplicative group of a field

Easy facts

 * Multiplicative group of a field implies every finite subgroup is cyclic
 * Multiplicative group of a finite field is cyclic, and in particular multiplicative group of a prime field is cyclic

Closed normal subgroups and quotients
For a finite field, the group is cyclic of order one less than the size of the field, and the topology is discrete, so the study proceeds just like the study of the subgroup structure of finite cyclic groups.

For an infinite field, the closed normal subgroups are precisely the finite subgroups and the whole group. Further, the finite subgroups are precisely the subgroups of $$n^{th}$$ roots of unity for positive integers $$n$$. Any such subgroup is cyclic of order dividing $$n$$ (and equal to $$n$$ if a primitive $$n^{th}$$ root exists, which is always the case for an algebraically closed field whose characteristic does not divide $$n$$).

The quotient group by this subgroup is isomorphic to the subgroup of the multiplicative group comprising all $$n^{th}$$ powers. This is because the closed normal subgroup of $$n^{th}$$ roots of unity is an endomorphism kernel corresponding to the endomorphism $$x \mapsto x^n$$. In particular, when the field is an algebraically closed field, the endomorphism is a surjective endomorphism and thus the quotient by the subgroup of $$n^{th}$$ roots of unity is isomorphic to the multiplicative group itself.