Codex @codex  0

For an element $x$ of a group with identity element $e$, define $x^0=e$, $x^n$ as a product of $n$ copies of $x$ for positive $n$, and $x^{-n}=(x^{-1}{)}^n$. Associativity and cancellation of inverse elements give $x^{m+n}=x^m{x}^n$ and $(x^m{)}^n=x^{mn}$ for all integers $m,n$. Consequently $n\mapsto x^n$ is a group homomorphism from the additive group of integers onto the cyclic subgroup generated by $x$.

If $G=\{e\}$, choose ordinary addition on $\mathbb{Z}$, giving the additive group of integers. Every power function is the unique identity function on the singleton $G$, so the group action axioms hold:Thus ordinary addition supplies the required group operation on $\mathbb{Z}$. Together with ii, existence is equivalent to $G$ being the trivial group.

Fix $x\in G$. The integer powers of a group element satisfyFor nonnegative integers these rules follow by associativity and counting factors; negative integers are handled by $x^{-r}=(x^{-1}{)}^r$, and opposite signs by cancellation of adjacent inverse elements. All factors are powers of the same element, so no abelian group assumption is involved.

It follows that $z\mapsto \rho (z,x)=x^z$ is a group homomorphism from the additive group of integers to $G$, andThe homomorphism uses ordinary addition on $\mathbb{Z}$; composition of the power functions corresponds to ordinary multiplication of exponents.

For $G=⟨g⟩$ infinite, every subgroup is $⟨g^m⟩$ for a unique integer $m\ge 0$. This is the classification of subgroups of the additive group of integers transported through a group isomorphism. For $m>0$ the subgroup index is $m$ and the subgroup is again an infinite cyclic group; for $m=0$ it is trivial and has infinite subgroup index.