TITLE: Semi group of polynomials which all roots lie on the unit circle
QUESTION [9 upvotes]: Let $X=\{f\in \mathbb{C}[z]\mid |z| \neq 1 \implies f(z) \neq 0\} $.
The motivation for  consideration of such an $X$ is the the concept of Lee-Yang polynomials.
With the standard multiplication, $X$ is  an  Abelian semigroup with cancellation property.
Let $G$   be the Grothendieck group associated with $X$.
Is there  a well known group which is  isomorphic to $G$? In other words, is there  an alternative formulation  of $G$ in terms of some well known group? Is there a  natural topology  on $G$ which makes it a  locally compact topological group?

REPLY [6 votes]: Every locally compact group topology on $G$ makes the constants $\mathbf{C}^*$ an open subgroup. 
Indeed, as observed in Fetisov's answer, $G$ is a direct product $\mathbf{C}^*\times A$ with $A$ free abelian. So $\mathbf{C}^*$ is the intersection of all kernels of homomorphisms $G\to\mathbf{Z}$.
R. Alperin (1980) proved that every homomorphism from a locally compact group into $\mathbf{Z}$ is continuous. It follows that $\mathbf{C}^*$ is closed in $G$.
So, working in the quotient, it is enough to prove: the only locally compact group topology $T$ on a free abelian group $A$ is the discrete one. Indeed, every subgroup of $A$ is free abelian. So $(A,T)$ cannot have a non-trivial compact subgroup (e.g., using again Alperin's result). It follows (by Hilbert's fifth problem, but which was previously known, probably due to Pontryagin) that the zero component $(A,T)^\circ$ is isomorphic to $\mathbf{R}^k$, and again this forces $k=0$, so $A$ is discrete.
Now $\mathbf{C}^*$ admits plenty of exotic locally compact group topologies, but any reasonable assumption (e.g., $\sigma$-compact + evaluation at some point is continuous) will force the topology to be the canonical one.