Determining elements of group from generating relations Is there an algorithm for determining all the elements fo a finite group from its generating relations?
For example, let group $G$ have the generating relations $p^3 = q^2 = (qp)^2 = 1$. I see that $p$, $p^2$, and $q$ will each be elements. But how (beyond basically trial and error) do I determine which compositions are also distinct elements of $G$?
 A: Given that you know the group is finite, you can use the Todd-Coxeter process to enumerate the cosets of the trivial subgroup.  That, effectively, gives you the elements of the group.
A good description can be found in Johnson.  In practice, one would use a computer implementation to do the work.
A: While the Todd-Coxeter process will work (in principle) for any finite group given by a finite presentation (given enough time and memory), for specific cases, such as the example you provided, there may be another approach, which is to try to find a normal form for the group elements.  In your example, this approach works quite nicely.
Notice that your relations $(pq)^2 = 1$ and $q^2 = 1$ can be interpreted as
$q^{-1}pq = p^{-1} = p^2,$
or
$$qp = p^2q.$$
This allows you to move powers of $p$ to the left of $q$ in any expression. For instance,
$$qp^2 = p^2qp = p^4q.$$
Moreover, any power $p^k$ of $p$ can be written as $p^0 = 1$, $p$ or $p^2$, since $p^3 = 1$. Likewise, a power of $q$ is either $1$ or $q = q^{-1}$. Thus, we get, for example, that $qp^2 = p^4q = pq$.  In general, any string of powers of $p$ and $q$ can be put into the form
$$p^iq^j, \;\;\;\text{with $0\leq i<3$ and $0\leq j<2$.}$$
Counting these shows that your group has at most $6$ elements.  Unless you have some reason to suspect the group might actually be smaller, you'd now try to show that all these elements are actually distinct.  The standard way to do this is to find a homomorphism from your group onto some known group with $6$ elements.  Since the defining relations suggest that the group is not commutative, the only candidate is the symmetric group $S_3$.  And, in fact, the mapping
$$p\mapsto (1,2,3)\;\;\; q\mapsto (1,2)$$
is easily shown to define a homomorphism from your group onto $S_3$, because the images $(1,2,3)$ and $(1,2)$ of $p$ and $q$ satisfy the relations defining your group.  Having a homomorphism from your group onto $S_3$ shows that it has at least $6$ elements which, combined with the previous result, demonstrates that it has exactly six members.  (In fact, it shows that your group is isomorphic to $S_3$.)
