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.
 A: 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.
A: 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.
