Given a set $S$ in a group $G$, how does the smallest normal subgroup containing $S$ look like? In the theory of rings we have clear descriptions of the smallest ideal in a ring $R$ containing a subset $S$ of $R$. I'd like to know if there is such a description in group theory, that is, if $N(S)$ is the smallest normal subgroup of $G$ containing $S$, can you describe $N(S)$ using the operations of groups and the elements of both $S$ and $G$? Is there such description if $S$ is a subgroup? Thanks.
The smallset subgroup of $G$ containing the set $S$ is simply $\langle S \rangle$, the subgroup generated by all elements of $S$, as any smaller subgroup containing $S$ would not be closed.

To describe the smallest normal subgroup containing $S$, we do something similar.  We want for $s^g$ to remain in the group for any $s\in S, g\in G$, so we define $S^G=\{s^g:s\in S, g\in G\}$.  Then $\langle S^G \rangle$ is precisely what we want: the smallest subgroup of $G$ containing $S$ and all conjugates thereof.  As you can see there is no difference in these definitions if $S$ is a subgroup.

This actually has a name: $\langle S^G \rangle$ is the normal closure, or conjugate closure, of $S$ in $G$.  A bit more information on normal closures can be found here, though not much can be said about them in general without looking at specific cases.
It's going to be very complicated, nowhere near as simple as the description of the ideal generated by elements of a ring.  For example, in the free group on 2 letters $a$ and $b$, the smallest normal subgroup containing the single element $aba^{-1}b^{-1}$ consists of all words where the signed total number of $a$'s (i.e. counting $a^{-1}$ as negative) equals the signed total number of $b$'s.  And this is a relatively simple case to determine.  In general, there exist group presentations having unsolvable word problem, i.e., there is no algorithm for determining whether a particular word belongs to the normal subgroup generated by the relations, and those cases will be extremely complicated.