I had a feeling that the multiplicative group of the complex numbers C∗\mathbb{C}^* isn’t isomorphic to the circle group S1S^1, just because S1S^1 embeds into C∗\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.

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The group ZN\mathbb{Z}^{\mathbb{N}} (direct product) embeds in ZZ\mathbb{Z}^{\mathbb{Z}}. However, they’re isomorphic.

– egreg

2 days ago

Right, or Z\mathbb{Z} and 7Z7\mathbb{Z} will do the same trick. So embedding won’t give an argument, but I was looking for something else.

– ougoah

2 days ago

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2 Answers

2

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For both groups, the torsion part is isomorphic to Q/Z\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 Q\mathbb{Q}-vector spaces.

So the circle group and C∗\mathbb{C}^* are indeed isomorphic.

This is the first I’ve heard about divisible groups and their structures. Looks like interesting stuff!

– ougoah

2 days ago

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Note the essential use of the Axiom of Choice. You cannot constructively “write down” such an isomorphism! Same problem as with R\mathbb R and R2\mathbb R^2 additive groups.

– GEdgar

2 days ago

1

@ougoah There’s not much to say: a divisible abelian group is of the form Q(α)⊕⨁pZ(p∞)(βp)\mathbb{Q}^{(\alpha)}\oplus\bigoplus_p \mathbb{Z}(p^\infty)^{(\beta_p)}, where Z(p∞)\mathbb{Z}(p^\infty) is the Prüfer pp-group (the sum runs over the primes) and α,βp\alpha,\beta_p are cardinal numbers uniquely determined. The most interesting aspect is the proof of this.

– egreg

2 days ago

The above statement can be answered from representation theory point of view as follows.

Assume that by T\mathbb{T} I denote the circle group i.e. the set T={z∈C∣|z|=1}\mathbb{T}=\{ z \in \mathbb{C} \mid |z|=1 \}. Furthermore, assume that ρ:T→C∗ \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 C∗\mathbb{C}^{*} isn’t, such a homomorphism cannot be isomorphism! In fact every such a homorphism is an endomorphism of T\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 R⊕Q/Z\mathbb{R} \oplus \mathbb{Q}/ \mathbb{Z}. But sometimes topology makes things rather more complicated and needs a bit more attention!

But disregarding topology, is it still true that they are not isomorphic as (abstract, not topological) groups?

– ougoah

2 days ago

I just tried to give you an alternative about how the things are a bit complicated sometimes to be honest. if you forget the topological structure then they are isomorphic

– mayer_vietoris

2 days ago