Smallest subgroup containing a subset $A$ Let $G$ be a group and let $A \subset G$. I want to show that there is a $\subset$-least subgroup $H$ of $G$ such that $A \subset H$. The subgroup is noted as $\langle A\rangle$.
Then I want to show that $\langle A\rangle$ is the set of all finite products of the elements of $A$. 
Kinda confused on how to start this one. I know that $\langle A\rangle$ us a subgroup of $G$, so it is closed under multiplication and inverses and also contains $A$. Therefore, $\langle A\rangle$ must contain the products that are in $H$?
 A: Normally the way these sorts of arguments work is just deciphering what is meant by "least". In this case, one can show that the intersection (over an arbitrary index set) of subgroups will also be a subgroup. Then $\langle A \rangle$ is clearly going to be just this intersection (understand this). Of course, one must also know that the index set we are intersecting over is non-empty: can you think of a subgroup of $G$ containing $A$?
Once you've established this formalism, the description of $\langle A \rangle$ as the set of all finite products of the elements (and their inverses) is easy to see. Show that this set is a group, and then show that any subgroup of $G$ must contain these elements.
A: The set of all finite products will not give a subgroup, for example taking $G:=\mathbb Z$ with usual addition and $A:=\{2\}$. Then the set of all finite products will be $2\mathbb N$ which is not a subgroup. 
However, if we define $A^{—1}$ as the set of the inverse of elements of $A$, then one can show that the collection of finite products of elements of $A\cup A^{-1}$ is the subgroup generated by $A$. 
A: Your starting is good: it shows that $<A>$ must contain all finite products of elements of $A$ and their inverses, because it is a group containing $A$.
For the converse, look at the object "set of finite products of elements in $A$ and their inverses", and try to prove that it has to contain $<A>$, by definition of $<A>$ as minimal subgroup containing $A$.
