Generating set of a group A subset $S$ of a group $G$ is said to be a generating set for $G$ if all elements of G can be expressed as the FINITE product of elements in S and their inverses. Why is it necessary to take only finite expressions ? Is there a group $G$ and a set $S$, where not all elements of $G$ can be written as a finite expression taken from the set $S$ ?
 A: A group is a set $G$ together with binary operation that is a map from $G\times G $ to $G$ satisfying some set of axioms. It means that you are allowed to apply this operation to the pairs of elements only. The product of arbitrary finite number of elements is defined by induction: $a_1 \dots a_{n-1}a_n = (a_1 \dots a_{n-1})a_n$. If you want to form infinite (e.g. countable) products of elements you should define a map from the direct power $G^{\mathbb{N}}$ (the set of all functions $f\colon \mathbb{N} \to G$) to $G$, which can be done in many different ways. The main point here is that operation of finite product can be represented (to be written as a term) using only binary product, while infinitary one can not. With this additional operation you get the new structure, that is a group endowed with infinitary operation. 
The definition of a subgroup in this structure differs from the usual one. A subgroup of a group $G$ is a subset of $G$ closed under the product operation (which is equivalent to be closed under arbitrary finite products) and taking the inverse element. While the subgroup of a group endowed with infinitary operation should be closed under this infinitary operation too. In order to see the difference look at Darth Geek's example in comments.
For the second question take $G = \mathbb{Z}$ with usual addition operation and $S = \{2\}$. You can't write $1$ as a finite expression of elements of $S$. Taking all possible finite expressions you will get the set of all even numbers $2\mathbb{Z}$, it is the subgroup generated by $S$.
