Cyclic implies every field is class-determining

Statement

If $G$ is a cyclic group and $K$ is a field, then $K$ is a class-determining field for $G$, i.e., given two representations $\varphi _{1},\varphi _{2}:G\to GL(n,K)$ with the property that $\varphi _{1}(g)$ and $\varphi _{2}(g)$ are conjugate for every $g\in G$, it is true that $\varphi _{1}$ and $\varphi _{2}$ are equivalent linear representations, i.e., there exists a matrix $A\in GL(n,K)$ such that $A\varphi _{1}(g)A^{-1}=\varphi _{2}(g)$ for all $g\in G$.

Related facts

• Elementary abelian of prime-square order implies corresponding prime field is not class-determining

Proof

The key idea is that we find the matrix that works for the generator of the cyclic group, and show that it works for the whole group.