Can there exist a group isomorphism between $\mathbb{C}^*$ and $S^1$? I had a feeling that the multiplicative group of the complex numbers $\mathbb{C}^*$ isn't isomorphic to the circle group $S^1$, just because $S^1$ embeds into $\mathbb{C}^*$ (which isn't an argument in itself, as egreg points out in the comments below). I'm looking for a proof of this fact (if it indeed is a fact).
Any thoughts would be great.
 A: For both groups, the torsion part is isomorphic to $\mathbb{Q}/\mathbb{Z}$. Since the groups are divisible, the torsion part factors out. The quotients modulo the torsion part are torsion free divisible groups with the same cardinality, hence isomorphic as $\mathbb{Q}$-vector spaces.
So the circle group and $\mathbb{C}^*$ are indeed isomorphic.
A: The above statement can be answered from representation theory point of view as follows.
Assume that by $\mathbb{T}$ I denote the circle group i.e. the set $\mathbb{T}=\{ z \in \mathbb{C} \mid |z|=1 \}$. Furthermore, assume that $ \rho : \mathbb{T} \rightarrow \mathbb{C}^{*}$ is a 1-dimensional representation (which in fact is a homomorphism of groups in that case), by Shur's Lemma now, we know that all those maps are given irreducible representations and it's not difficult to conclude that the above map is in fact faithful (is another variance of Schur's Lemma actually). Since that group is compact and $\mathbb{C}^{*}$ isn't, such a homomorphism cannot be isomorphism! In fact every such a homorphism is an endomorphism of $\mathbb{T}$.
I will edit my answer because I think there is a misunderstood regarding my answer (maybe correct). They are indeed isomorphic, since they are both isomorphic to  $\mathbb{R} \oplus \mathbb{Q}/ \mathbb{Z}$. But sometimes topology makes things rather more complicated and needs a bit more attention!
