Identify the group generated by two elements with given relations Let us consider the group $G=<a,b| a^3=b^2=e>$. 
Progress: The elements of the group are $e,a,a^2,b,ab,a^2b,ba,b^2a$  
How can I identify this group with a family of known group, like symmetric group or dihedral group etc.?  
 A: The canonical answer is hard to find if you don't already know it : this group is isomorphic to $PSL_2(\mathbb{Z})$, with the generators $a= \begin{pmatrix} -1 & 1 \\ -1 & 0\end{pmatrix}$ and $b = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$.
Of course you have to check that they don't satisfy any additional relation, which is not trivial : it can be shown geometically using the action by homography on the hyperbolic half-plane, and the so-called ping-pong lemma.
A: In general, it is hard to deduce group properties from a presentation. A prototipical example is that you can not algorithmically decide whether a given finite presentation presents the trivial group or not.
However, in some cases the situation is much more friendly. For example, it is easy to see that the free product of the groups with presentations $\langle X \mid R \rangle$ and $\langle Y \mid S \rangle$ admits a presentation having as generators the disjoint union of the generators of the factors, and having as relators the disjoint union of realators of the factors; i.e.,
\begin{equation}
\langle X \mid R \rangle *\langle Y \mid S \rangle = \langle X \sqcup Y \mid R \sqcup S \rangle \, .
\end{equation}
Following this scheme it is clear that the free product of the cyclic groups of orders two and three admits the given presentation:
\begin{equation}
C_3 * C_2 = \langle a \mid a^3=e \rangle * \langle b \mid b^2=e \rangle = \langle a,b \mid a^3=b^2=e \rangle \, .
\end{equation}
