Free $C^*$-Algebras I am confused by a passage in the wikipedia article Universal $C^*$-algebra, particularly these two sentences

This means that depending on the generators and relations, a universal C*-algebra may not exist. In particular, free C*-algebras do not exist.

My confusion arises in connection with one of the examples offered in the wikipedia article: 

The universal C*-algebra generated by a unitary element u has presentation $ \langle u | u^*u = uu^* = 1 \rangle$. By the functional calculus, this C*-algebra is the continuous functions on the unit circle in the complex plane. Any C*-algebra generated by a unitary element is the homomorphic image of this universal C*-algebra

Wouldn't a group generated by a single unitary element be a free group on one generator? If so, how can the article claim there are no free $C^*$-algebras? Perhaps I am misunderstanding their use of the word 'free'. In addition to resolving this confusion, I am also interested in knowing whether there are $C^*$-algebras which admit of unitaries that form a free group? Are they numerous? I am mostly interested in a finite number of unitaries forming a free group, if that makes things any easier. I think one way of doing this would be to take a finite collection of unitaries each having infinite order, and then take the free product of their cyclic groups; but I might be mistaken. Is there any other way finitely generated free groups of unitaries arise? 
 A: A free object to a set $X$ in a representable category is an object $A$ together with an inclusion $i:X\to F(A)$ so that for any other map $f:X\to F(B)$ there is a unique morphism $g: A\to B$ so that $F(g)\circ i = f$. (Here $F$ is supposed to be the forgetful functor.)
Notice how all you have is a set of elements, and no relations on these, unlike your example.
For example let $X$ be a two-point set, and suppose you have a free algebra $A$ wrt $X$ with $a_1=i(p_1)\neq i(p_2)=a_2$. You can define a map $f: X\to F(A)$ via $f(p_i)=2a_i$ and then you must have a morphism $g:A\to A$ with $g(a_i)=2a_i$. Now $C^*$ algebra morphisms are contractive, which is incompatible with $g$ unless both $a_1=0=a_2$, which is not allowed.
As to your second question, finding a $C^*$ algebra with unitaries coming from a group $G$, namely $G$ acts via unitaries on $\ell^2(G)$:
$$u_g\left(\sum_{h\in G} x_h\,h\right) := \sum_{h\in G} x_h gh.$$
You can verify $u_gu_h=u_{gh}$ and $u_e=\mathbb 1$. Taking the sub-algebra of $C(\ell^2(G))$ generated by the $u_g$ you get a $C^*$ algebra, but it can have a lot more unitaries than just those of the form $u_g$.
A: The C$^*$-algebra generated by a unitary $u$ has two relations, namely $u^*u=uu^*=I$, so it is not free. 
Given any free group, you can certainly generate a C$^*$-algebra with it. More than one, in fact. For a discrete group $G$ two canonical C$^*$-algebras that one can generate are the full C$^*$-algebra $C^*(G)$ (generated by the universal representation) and the reduced C$^*$-algebra $C^*_r(G)$, which is generated by the left regular representation. 
These two C$^*$-algebras are equal precisely when $G$ is amenable. So in particular, for the free groups $\mathbb F_n$, $n\geq2$, they are different. But $C^*_r(\mathbb F_2)$, say, is not free as a C$^*$-algebra, since the generator is taken as a unitary and thus has relations. 
Answering your other question: there are many ways in which some unitaries in a C$^*$-algebra can be free. 
