I would like to determine the order of the group $G = \langle x, y \mid x^3 = y^2 = (xy)^3 = 1\rangle$.
Here is a proof that $G$ contains at most $12$ elements. Any element $g \in G$ can be represented by a word in $x, y$. Using the relations $x^3 = y^2 = 1$, we can ensure that this word does not contain $x^3$ or $y^2$ as subwords. If $y$ still appears multiple times in the word, then there must be a subword of the form $yxy$ or $yx^2y$. By repeatedly applying the relations $yxy = x^2yx^2$ and $yx^2y = xyx$ (both of which drop the number of $y$'s by one), we obtain a word where $y$ appears at most once. The resulting word must be one of the twelve possibilities: $$ 1, x, x^2, y, yx, yx^2, xy, xyx, xyx^2, x^2y, x^2yx, x^2yx^2, $$ which proves $|G| \leq 12$.
However, I'm not sure how to prove that $G$ contains exactly twelve elements. I suppose I could give the above set of twelve elements a group operation, and show that it is well-defined, but that seems rather tedious. Another way might be to show that $G$ is equal to some familiar group, perhaps the automorphisms of a certain geometric space. I believe I can do this (by finding the Cayley graph of $G$), but I'm wondering if there is a more algebraic solution to this problem.