Upper bound for order of finite group given relations Say I have a group with the following presentation:
$$
G = \langle a,b \mid a^2 = b^3 = (ab)^3 = e \rangle
$$
During a conversation someone had mentioned that the order for $G$ must be less than or equal to $12$. I couldn't follow the conversation very well, but on trying to figure out where this bound came from I got confused. They seemed to make it sound like there was some certain property that allowed them to calculate this fairly rapidly. Is there some theorem that gives an upper bound to finite groups that are relatively nicely behaved? (Like those with two or maybe three generators).
 A: I don't think there is any general theorem of this sort, unless we know something reasonably specific about the relations.  For example, every finite simple group can be generated by two elements, albeit possibly with many relations.  Also disturbing:  the word problem for finitely presented groups is not solvable in general; that is, there is no algorithm that can determine whether two elements in an arbitrary finitely presented group are equal.  And finally, an open problem:  it is unknown whether a 2-generated group where all fifth powers are trivial must be finite.  (See $B(2, 5)$.)
A: I would go the most basic way : enumerate the elements.
So from $a^2 = b^3 = e$, you know that the elements of the group are sequences of $a$ and $b$ where there is at most one consecutive $a$, and two consecutive $b$.
Let us list the elements by their number of factors.


*

*0 factors : $e$

*1 factor : $a$ and $b$

*2 factors : $ab, ba$ and $b^2$, all distincts

*3 factors : $aba, bab, ab^2, b^2a$, all distincts

*4 factors : $abab, baba, ab^2a, b^2ab, bab^2$


But we have $baba = (ba)^{-1} = ab^2,\quad abab = (ab)^{-1} = b^2a$, and $ab^2a = (aba)^{-1} = bab$, so the only new element with 4 factors are $b^2ab$ and $bab^2$, whoch makes us 12 elements in total.
Now let's look at the elements with 5 factors : 
$ababa, abab^2, ab^2ab, babab, bab^2a, b^2aba, b^2ab^2$. Each one but the last contains a 4-factor that reduced to a 3-factor, so these are in fact 4-factors and we have already seen them. Now $b^2ab^2 = (bab)^{-1} = aba$ is a 3-factor.
Any factor of superior length would also be reducible, so we have enumerated all the elements, and there are at most 12 of them.
A: There are already several good answers, but I would like to mention one additional tool, because it is systematic: the Todd-Coxeter algorithm. Given a presentation of a group that includes a finite-order generator, if the group is finite, the Todd-Coxeter algorithm will terminate and provide the group order. (If there is a danger that the group is not finite, then you run the risk of the algorithm not terminating.)
I learned the algorithm from Chapter 6 of this book.
A: If you are able to give an upper bound for the order of a group, it's usually because you recognize $G$ as a quotient (homomorphic image) of a known finite group, say $\tilde{G}$.  In this case, you might notice that the tetrahedral group $\tilde{G} = A_4 < S_4$ can be given by the presentation
$$
\langle x, y \mid x^2 = y^3 = (xy)^3 = 1 \rangle,
$$
where $x = (1\,2)(3\,4)$ and $y = (1\,2\,3)$.
You check that the map $\varphi: A_4 \to G$, defined by
\begin{align}
x \mapsto a \\
y \mapsto b
\end{align}
is, in fact, a homomorphism.  This amounts to checking that for each relation in the variables $x$ and $y$, the corresponding relation is satisfied in the image of $\varphi$ in the variables $a$ and $b$.  (In this example, it's trivial because the map is so simple.)
Finally, you check that the map is onto.  (Again, this is trivial in this case.)
Now, you can conclude that $\lvert G \rvert \le \lvert \tilde{G} \rvert < \infty$.  You actually get a stronger conclusion:  by Lagrange's theorem, $\lvert G \rvert$ is a divisor of $\lvert \tilde{G} \rvert$.
