tl;dr: What is a minimal definition of a finite cyclic group, without resorting to the definition of a group?
Let $G$ be a finite set and $\star$ be a binary operation on $G$. My question is whether a cyclicity condition can imply some of the standard group axioms (as it implies for instance abelianity). Let us take a following cyclicity axiom:
$$\exists g\in G, \forall x\in G, \exists n\in ℤ_{>0}, x = g^n$$ where $g^n = g\star g\star\dotsb\star g$ ($n$ times). Already, this cyclicity condition assumes associativity to make sense of $g\star\dotsb\star g$. This implies that $g^{m+n} = g^m\star g^n$ for $m$, $n\inℤ_{>0}$. Note that I do not define $g^n$ for $n = 0$ or $n < 0$ that would assume identity or the existence of inverses.
Some thoughts about the other (standard) axioms:
- Closure: It seems required to me, as some kind of a converse to cyclicity (for all $n >0$, $g^n\in G$). This could be paired with cyclicity by writing something like: "a finite set $G$ is a cyclic group if $G = \{g^n:n\inℤ_{>0}\}$ for some $g\in G$."
- Identity: Since $G$ is finite, there must exist $m < n$ such that $g^m = g^n$. With only the other axioms, can we prove that there exists $k$ such that $g^k = id$ (with standard definition for identity)?
- Inverses: If we assume identity, we can give a sense to $g^0$ and prove that for all $x\in G$, there exists $0 ≤ n < k$ such that $x = g^n$. Then assuming closure, $g^{k-n}\in G$ and $g^{k-n}$ is the inverse of $x$.
My impression is that closure is required, inverses are not and I do not know for identity. But there may exist other ways to define finite cyclic groups.
You could take Peano's axioms and replace the axiom which states $0$ is not the successor of any natural number with its negation (number 2 below).
Your axioms then are, $G$ is a set together with a map $S:G\to G$ ('successor'), and an element $z\in G$ ('zero') such that
Once you have a model $(G,S,0)$ for these axioms, define the group law by:
which recursively defines addition.
*From the comments, it seems clear that I should explain why all models are finite cyclic groups. First note that $\{z,S(z),S^2(z),...\}$ must be all of $G$ by 3. Moreover, as 2 says that $z\in \operatorname{im}(S)$, we can conclude that $z=S^n(z)$ for some $n\in \mathbb{N}\setminus\{0\}$. Choose the least $n\in \mathbb{N}\setminus \{0\}$ such that $S^n(z)=z$. Another use of 3 implies that $\{z,S(z),S^2(z),...,S^{n-1}(z)\}$ is all of $G$, and minimality implies these are distinct.