I am trying to understand how (and why) the representation of a group $G$ and its (universal) $C^*$-algebra $C^*(G)$ entail the same information (just to let you know where the motivation comes from).
I will assume throughout that the group $G$ is countable.
To my understanding, a representation of a group is a group homomorphism from $G$ to some $B(\mathcal{H})$, in the case of a $C^*$ algebra it is a norm preserving $*$-homomorphism onto $B(\mathcal{H})$. Since a one-dimensional space is canonically isomorphic to $\mathbb{C}$ and $B(\mathbb{C})\cong \mathbb{C}^{\times}$ (multiplication without zero); what we really want is a map $$t:G\to \mathbb{C}^{\times}$$ and/or $$t:C^*(G)\to \mathbb{C}^{\times}$$ that are the appropriate kinds of morphisms (respectively).
Now Since $t:G\to\mathbb{C}^{\times}$ is a group homomorphism so $t(g^{-1}g)=1=t(g^{-1})t{g}$ but if $G$ is finite that means that $|t(g)|=1$ otherwise the norm of $|t(g^n)|$ would increase, so all $t(g)$ must be unitary. So for any finite group $t(G)\subseteq S^1$. (this is clearly compatible with the representation of $C^*(G)$ since we need to have $t(u_g^*)=t(u_{g{^-1}})=t(u_g)^*$; which means, in the group setting that $t(g^{-1})=t(g)^*$). Now my first question is, what is the image of $F_2$ the free group under some (nontrivial) one-dimensional representation? What can we say about $t(F_2)\subseteq\mathbb{C}^\times$, what is the maximal image this representation can give, is it still going to be contained in $S^1$ for example?
If we write $F_2$ as a free product of two copies of the infinite cyclic group, $F_2 = \langle a \rangle * \langle b \rangle$ then a homomorphism $t : F_2 \to \mathbb C^\times$ is uniquely and freely determined by choosing $t(a), t(b) \in \mathbb C^\times$. Any choices at all are possible, and the image $t(F_2)$ will then be the subgroup of $\mathbb C^\times$ generated by $t(a)$ and $t(b)$. So in particular you can choose $t(a)$ and $t(b)$ to be not contained in $S^1$, which gives plenty of examples where $t(F_2) \not\subset S^1$.
One thing that's happening here is that because $\mathbb C^\times$ is abelian, any homomorphism $t : F_2 \to \mathbb C^\times$ factors through the abelianization map $F_2 \mapsto \mathbb Z^2$, and therefore $t$ is uniquely and freely determined by the choice of homomorphism $\mathbb Z^2 \to \mathbb C^\times$.