Free group presentation

Question

So a free group is just a set $G$ with relations corresponding to the group axioms. Normally these relations are omitted from the presentation of a free group. The identity and inverse are easy but how do we write the relation for associativity? For example, say our set $X$ has order 4 then $F(X)$ is (please fill in the last relation "?????" corresponding to associativity): $$ <e,a,b,c| ea=a, eb=b, ec=c, ab=e, c^2=e, ????> $$ It's probably something like $(ab)c=a(bc)$.

Answer 1

The $\langle\,\ldots\mid\ldots\, \rangle $ notation by definition stands for a group that is generated by the elements on the left and such that the relations on the right hold (and simply put, for "the most general" group with these restrictions). Thus the cyclic group of order $4$ can be given by the representation $\langle\,a\mid aaaa=1\,\rangle$. Let's play along and say that the notation rather should stand for a magma. In that case, it would be simplest to list not just a small set of generators on the left but rather all elements of the group, and on the right to spell out the full multiplication table, such as $$\langle\,e,a,b,c\mid\,ee=e,ea=ae=a,eb=be=b,ec=ce=c,ab=ba=cc=e,ac=ca=b,bc=cb=a,aa=bb=c\rangle $$ We can try to save a few symbols on the left, i.e., get closer to the situation of listing just generators: $$\langle\,e,a\mid ee=e,ea=ae=a,(aa)e=e(aa), (aa)a=a(aa), (aa)(aa)=((aa)a)a=e\,\rangle $$ but I'm afraid I forgot some: Can we really conclude from those relations that $((aa)a)((aa)a)=aa$? Actually, we can't. Also note that we must keep the $e$ in the list of generator or replace it with a cumbersome expression such as $((aa)a)a$ throughout. At any rate, this method ends up as a nightmare of impossibilities once we want to represent infinite groups this way. Main part of the nightmare is that the corresponding concept to normal subgroup in the category of magmas is not as nice.

Then how do we know that this (i.e., the first variant) is associative? Well, we can check that for every instance it follows that the corresponding equality holds. for example(!) we have $(ab)c=a(bc)$ because per the relations $(ab)c=ec=c$ and $a(bc)=aa=e$. But in a more general situation this would not be sufficient, the main reason being that associativity among generators does not imply associativity in general!

So we see that there is good reason that the group theoretical notation used for representing groups actually uses some group theory ...