Why does the free group of the singleton set contain all powers of the group element? I'm following Paolo Aluffi's Algebgra 0, and in II.5.5.1 it talks about free groups.
It comes to say that the free group of the singleton set $\{a\}$ is the set of all powers of $a$, identifying $a^0 = e$ where $e$ is the group identity.
I don't understand the intuition behind this definition. Why isn't the free group the group with 2 elements, $e$ and $a$, and such that $a*a=e$?
Note that I understand that free group is given by a universal property of commutativity of diagrams, but the intuition behind this seems to be that it captures $\{a\}$ in the most efficient way. So my question is simply why isn't the group with only two elements, if valid, considered a simpler encoding? Why are all the powers needed?
EDIT: Clarification
 A: "Free" essentially means with as few assumptions as possible. So the the free group $F_S$ generated by a set $S$ must contain $S$ as a subset, and by group axioms must contain all "words" comprised of "letters" of $S$ as well as their inverses (which must exist). However, the whole point of "freeness" is to be as permissible as possible, or in other words as minimally restrictive as possible, or in other words as un-presumptuous as possible about its elements, which means we can't make any assumptions about group elements unless they follow logically from the group axioms. We cannot assume $a^2=e$.
(The point of "freeness" is not to be as "efficient" as possible, which I interpret you to mean as small as possible provided the set $S$ is a minimal generating set. In this case, the most "efficient," or I would simply say smallest, group containing $S$ as a minimal generating set would be the elementary abelian group $\mathbb{Z}_2^S$.)
Another way of thinking about the free group is that it is the most universal group containing $S$. That is, if there is any group $G$, and we have any function $f:S\to G$, then it extends uniquely to a group homomorphism $F_S\to G$. In the event $S=\{a\}$, we can't assume $a^2=e$ because then we couldn't be guaranteed the existence of a homomorphism $F_S\to \mathbb{Z}_3$ in which $a\mapsto 1$ (you can't have an element of order two map to an element of order three).
If $F_S$ is the free group on one element, then there must exist a one-to-one correspondence between functions $\{a\}\to G$ and group homomorphisms $F_S\to G$, and functions $\{a\}\to G$ are already in one-to-one correspondence with elements $G$. The range of the induced homomorphism $F_S\to G$ must contain an element of $G$ and hence the cyclic subgroup of $G$. By the first isomorphism theorem, the range is isomorphic to $F_S$ mod the kernel, so every cyclic group must be a quotient of $F_S$. We cannot assume $a$ has any finite order, because that would preclude us from extending a function $f:S\to G$ which maps $a$ to an element of order coprime to $n$ to a group homomorphism $F_S\to G$. Thus, $a$ has infinite order.
(This relates to what bof said in the comments.)
I also recommend looking into group presentations (as Peter Taylor mentions in the comments). In general, if $S$ is any set and $R$ is a set of "relations" then $\langle S\mid R\rangle$ is the "most universal" of "free-est" group generated by $S$ subject to the restraint that the elements of $S$ satisfy the relations specified by $R$. For example, we can get the cyclic group $\mathbb{Z}_n$ as $\langle a\mid a^n=e\rangle$ (often the "$=e$" part of the relations is suppressed). Or, we can get the dihedral group $D_n$ of rank $n$ (and order $2n$) as $\langle r,f\mid r^n,f^2,frf^{-1}=r^{-1}\rangle$ because for a minimal rotation $r$ and flip $f$ within this group, all of the relations between $r$ and $f$ can be deduced from three: that applying $r$ $n$-times returns to normal, that $f$ is a flip (i.e. involution), and conjugating a rotation by a reflection yields its inverse (which is true of every rotation and reflection in the plane).
Note that if $R$ has "more" relations than $R'$, then $\langle S,R\rangle$ will be a quotient of $\langle S,R'\rangle$, as making more assumptions about the group elements is more restrictive and yields a smaller group. If $R$ is the "empty set" of relations, so no assumptions are made at all about how the elements of the generating set $S$ interact, then we get the free group $F_S=\langle S\mid \varnothing\rangle$. This suggests $F_S$ is "as big" as possible containing $S$ (as a minimal generating set, anyway), which is in some sense the opposite as being the "most efficient" containing $S$.
Another illustration of this "free" idea is free products $G\ast H$, in which we want the "freest" group containing $G$ and $H$ as subgroups. It must contain all the elements of $G$ and $H$, and must contain all "words" of $G\sqcup H$, but we cannot assume anything that allows us to simplify these words except the operations already given in $G$ and $H$. In terms of presentations then it becomes intuitive that $\langle S_1,R_1\rangle \ast \langle S_2,R_2\rangle \cong \langle S_1\sqcup S_2,R_1\sqcup R_2\rangle$ for example.
