Additive characters are exactly one-dimensional complex representation characters

For $g\in G$ define ${\rho }_{\chi }(g):\mathbb{C}\to \mathbb{C}$ by ${\rho }_{\chi }(g)(z):=\chi (g)z$. This map is $\mathbb{C}$-linear and invertible with inverse $z\mapsto \chi (g{)}^{-1}z$, because $\chi (g)\in {\mathbb{C}}^{\times }$; moreover ${\rho }_{\chi }(0)$ is the identity map, since $\chi (0)=1$, and ${\rho }_{\chi }(g+h)={\rho }_{\chi }(g)\circ {\rho }_{\chi }(h)$ for all $g,h$, because $\chi (g+h)=\chi (g)\chi (h)$ and multiplication in $\mathbb{C}$ is associative. So ${\rho }_{\chi }$ is a group homomorphism $G\to \mathrm{GL}(\mathbb{C})$, hence a complex representation of $G$ on the one-dimensional space $\mathbb{C}$, and its defining formula in terms of $\chi$ determines it uniquely from $\chi$.

For clause 2 let $\rho :G\to \mathrm{GL}(V)$ be a one-dimensional complex representation, so ${\dim }_{\mathbb{C}}V=1$, and choose a basis vector $v$ of $V$. For each $g$ there is a unique scalar $\lambda (g)$ with $\rho (g)v=\lambda (g)v$, and $\rho (g)$ invertible forces $\lambda (g)\mathrm{\ne }0$; by [L9] the resulting map $\lambda :G\to {\mathbb{C}}^{\times }$ is a group homomorphism, hence an additive character of $G$. Since $\rho (g)$ is multiplication by the scalar $\lambda (g)$ on a one-dimensional space, its trace is $\lambda (g)$, so the trace character of $\rho$ is $\lambda$; and because the trace of an endomorphism is basis-independent, the additive character so obtained does not depend on the choice of basis.

Every irreducible complex representation of $G$ has degree $1$: by [L7] the fundamental theorem of algebra makes $\mathbb{C}$ an algebraically closed field, so [L8] makes every endomorphism of an irreducible complex representation $V$ of $G$ a scalar $\lambda {\mathrm{id}}_V$; since an irreducible $V$ is nonzero, the map $\lambda \mapsto \lambda {\mathrm{id}}_V$ is injective, hence ${\mathrm{End}}_G(V)=\mathbb{C}$ and $\mathbb{C}$ is a splitting field for $G$, whereupon [L8] gives ${\dim }_{\mathbb{C}}V=1$ because $G$ is abelian.

For clause 4 fix $g\in G$ and put $n:=\mathrm{ord}(g)$; since $G$ is finite, $n$ is a natural number with $n\ge 1$ and $g^n=e$. Written additively, as this page writes $G$, the power $g^n$ is the $n$-fold sum $n\cdot g$, so $n\cdot g=0$, and the additive reading of claim 3 of [L10] gives $\chi (n\cdot g)=\chi (g{)}^n$. Hence $1=\chi (0)=\chi (g{)}^n$: the value $\chi (g)$ is an $n$-th root of unity. By [L11] there is $k$ with $0\le k<n$ and $\chi (g)=\exp (2\pi ik\mathrm{/}n)$, and the modulus formula in [L11] gives $\mathrm{\mid }\chi (g)\mathrm{\mid }=\mathrm{\mid }\exp (i\cdot 2\pi k\mathrm{/}n)\mathrm{\mid }=1$.

In the standard basis $\{1\}$ of $\mathbb{C}$ the matrix of the endomorphism ${\rho }_{\chi }(g)$ is the $1\times 1$ matrix $(\chi (g))$, whose trace is $\chi (g)$; since the character is computed with the basis-independent trace, ${\chi }_{{\rho }_{\chi }}(g)=\mathrm{tr}({\rho }_{\chi }(g))=\chi (g)$ for every $g\in G$. Thus the trace character of ${\rho }_{\chi }$ is $\chi$, as clause 1 asserts.

Let $W$ be an irreducible complex representation of $G$. By step 1.3 it is one-dimensional, so by step 1.2 its trace character $\psi (g):=\mathrm{tr}({\rho }_W(g))$ is an additive character of $G$; by step 2.1 the representation ${\rho }_{\psi }$ attached to $\psi$ has trace character $\psi$. Thus $W$ and ${\rho }_{\psi }$ are degree-one representations producing the same homomorphism $\psi$, so [L9] makes them equivalent. This is clause 3.

For clause 5 let $\chi \mathrm{\ne }\psi$ be additive characters. Each ${\rho }_{\chi }$ is irreducible: it lives on the nonzero space $\mathbb{C}$ of dimension $1$, and a subrepresentation $U$ is a linear subspace with ${\dim }_{\mathbb{C}}U\le 1$, so either ${\dim }_{\mathbb{C}}U=0$ and $U=0$, or ${\dim }_{\mathbb{C}}U=1={\dim }_{\mathbb{C}}\mathbb{C}$ and $U=\mathbb{C}$ by [L5]; by [L4] this is exactly irreducibility, and by step 2.1 the character of ${\rho }_{\chi }$ is $\chi$, so ${\rho }_{\chi }$ has irreducible character $\chi$ in the sense of [L3]. If ${\rho }_{\chi }$ and ${\rho }_{\psi }$ were equivalent, then their characters would be equal by the invariance clause of [L3], so step 2.1 would give $\chi =\psi$, contrary to hypothesis; hence the two irreducible representations are inequivalent.

Clause 1 is steps 1.1 and 2.1, clause 2 is step 1.2, clause 3 is steps 1.3 and 3.1, clause 4 is step 1.4, and clause 5 is step 3.2, so all five clauses of the statement are proved. The argument uses only finite groups, finite-dimensional complex representations and finite lists of roots, and therefore invokes no choice principle.