I'm working out the exercises in MacLane and Birkhoff's Algebra. The exercise in question is the II.5.8:
where $\Delta_4$ stands for the fourth dihedral group.
Now I'm stuck trying to answer the first part because my reasoning leads to further equalities that nowhere help me build a Cayley table. I'm actually lost with the procedure itself. For now what I've be trying is to assume there are the elements $1,a,a^2,a^3$, but I can't find the remaining elements.
From the fact that $a^4 = 1$ I conclude $a^{-1} = a^3$, which in turn implies that $$b^{-1} a b = a^3$$
In the other hand, $a^2 = b^2$ leads to multiple new relations, which will further affect the consequences of $b^{-1}ab = a^{-1}$, for example concluding that $ba = aba^2$, which doesn't help.
This is, the defining relations are not helping me find which elements belong to the group (is $b^{-1}$ really in the group, or something like $ba^2$ takes its place?). In the case of $\Delta_3$ I was able to build its Cayley table from its defining relations because I knew beforehand which elements where in, but it is not the case here. I would really appreciate a complete explanation on how we build the Cayley table of a group, given a set of defining relations, but not the elements of the set.
Thank you in advance.
Let me try to frame the problem in another way. Start with all elements formed from the symbols in $\{a, a^{-1}, b, b^{-1}\}$, like $a^{10}$ or $b^3a^{-1}b^{-2}$. Now, with the relations given, you will notice that most of these are actually "the same" as each other (e.g. $a^3 = a^{-1}$, as you've pointed out).
As Derek Holt commented, what you want to do at this point is show that none of the elements of the form $a^ib^j$ with $i = 0, 1, 2, 3$ and $j = 0, 1$ are "the same" as each other (so you know you have at least 8 different elements in your group), and that every string made from the symbols in $\{a, a^{-1}, b, b^{-1}\}$ is "the same" as one of them (so you know you have exactly 8 elements in your group).
From this it will be easy to build your Cayley table.