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?



Semi group of polynomials which all roots lie on the unit circle
A complex polynomial is uniquely determined by its set of roots together with multiplicities. This means that the semigroup of your polynomials is freely generated by the set of point on the unit circle, aka by $\{z-k | k:\mathbb C, |k|=1\}$. Its Grothendieck group is the free abelian group on the continuum of generators. It can be given a nicer structure if you consider the natural topology on the set of polynomials.


In this case it will be the free topological abelian group generated by the unit circle $S^1$. This space is also not particularly nice, but up to homotopy equivalence it is $S^1 \times \mathbb Z$. This follows from Dold--Thom theorem which states that the free topological abelian group of a CW-complex $X$ is homotopy equivalent to the space that represents reduced singular homology $\tilde H\_\*(X)$.


The reduced homology of $S^1$ is $0, \mathbb Z, 0, \dots$, which corresponds to the homotopy groups of Eilenberg--Maclane space $S^1$. An extra factor of $\mathbb Z$ comes from connected components. Geometrically it means that any set of points on $S^1$ with multiplicities is equivalent to $n\cdot 1$, where $1: S^1$ is the basepoint. 


