Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 6 / b (When power maps can form an action)

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.

Suppose an arbitrary binary operation $\star$ makes $\mathbb{Z}$ a group and makes $\rho$ a left group action on $G$. For any group action, each acting element induces a bijection: the function induced by its inverse element is its inverse function.

In particular, the acting element whose underlying integer is $0$ must induce a bijection $G\to G$. But its power function isa constant function. This is injective only when all elements of $G$ equal $e$. ThereforeThis proves the power-map group action obstruction for every possible $\star$; it never assumes that $\star$ is ordinary integer multiplication, or that the integer $0$ is the acting identity element.

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.