The converse is true because diagonal matrices commute:
If you look at the $G$-module $\mathbb{C}[G]$, Maschke's theorem gives you a basis $\mathcal{B}$ for $\mathbb{C}[G]$ in which
$$\text{mat}_{\mathcal{B}}(\pi(g)) = \left(\begin{matrix}\chi_1(g) & &\\& \ddots&\\&&\chi_n(g)\end{matrix}\right)$$
Where $\chi_1,\ldots, \chi_n$ are characters of $G$ (where we denote $\pi(g) : \mathbb{C}[G] \to \mathbb{C}[G]$ the action of $g \in G$ on that space). Now, because the module is faithful, the morphism $g \mapsto \text{mat}_{\mathcal{B}}(\pi(g))$ from $G$ to $\text{GL}_n(\mathbb{C})$ is injective. But obviously the image of this map is a commutative group, so $G$ is also commutative.
The question of how to classify all finite-dimensional representations of $C_n$ over an arbitrary field $F$ can be studied using the structure theorem for finitely-generated modules over a principal ideal domain, in this case $F[x]$. The structure theorem asserts that any finitely-generated module is uniquely a finite direct sum of modules of the form $F[x]/p(x)^r$ where $p \in F[x]$ is irreducible and $r$ is a non-negative integer.
If $T$ is an operator acting on $F^k$ for some $n$, then $F^k$ becomes a finitely-generated module over $F[x]$ with $x$ acting by $T$. $T$ gives a representation of the cyclic group $C_n$ if and only if $T^n = 1$, in which case the summands $F[x]/q(x)^r$ in the decomposition of $F^k$ must have the property that $q(x)^r | x^n - 1$.
If $F$ has characteristic $0$ or has characteristic $p$ and $p \nmid n$, then $x^n - 1$ is separable over $F$, hence $r \le 1$ and $F^k$ is a direct sum of irreducible representations, all of which are of the form $F[T]/q(T)$ where $q$ is an irreducible factor of $x^n - 1$ over $F$.
If $F$ has characteristic $p$ and $p | n$, then writing $n = p^s m$ where $p \nmid m$ we have
$$x^n - 1 = (x^m - 1)^{p^s}$$
from which it follows that $r \le p^s$ (but now it is possible to have $r > 1$). If $r > 1$, then the corresponding representation $F[T]/q(T)^r$ is indecomposable and not irreducible, where $q$ is an irreducible factor of $x^m - 1$ over $F$. The irreducible representations occur precisely when $r = 1$. In other words,
The irreducible representations of $C_{p^s m}$, where $p \nmid m$, over a field of characteristic $p$ all factor through the quotient $C_{p^s m} \to C_m$.
One can also see this more directly as follows. If $V$ is an irreducible representation of $C_{p^s m}$ over a field of characteristic $p$ and $T : V \to V$ is the action of a generator, then
$$T^{p^s m} - 1 = (T^m - 1)^{p^s} = 0.$$
Thus $T^m - 1$ is an intertwining operator which is not invertible, so by Schur's lemma it is equal to zero.
Best Answer
In the case of an infinite topological group, you probably want to consider only continuous characters. In the compact case, the Peter-Weyl theorem holds, and the situation is similar to the finite case. If you consider arbitrary representations of $\mathbb{R}$, then you can construct some uninteresting and pathological ones from a Hamel basis. In the abelian case, the characters are more or less determined by Pontryagin duality, assuming local compactness. For the particular case of $\mathbb{R}$, it follows that the continuous characters on $\mathbb{R}$ are all of the form $\chi(t) = e^{2\pi i \xi t}$ for some real $\xi$, with the correspondence induced by the Fourier transform.