It is easy to show (using an argument involving $\lvert z\rvert$) that the finite subgroups of $\mathbb{C}^\times$ are exactly $\mathbb{Z}/n\mathbb{Z}$ for all $n\in \mathbb{N}$ embedded as the $n^{th}$ roots of unity.

A more interesting question is: can we classify all subgroups of $\mathbb{C}^\times$?

The finite case has been taken care of. Here is a tentative list:

$(1)$ The finite subgroups embedded as $\mathbb{Z}/n\mathbb{Z}$.

$(2)$ ${S}^1\hookrightarrow \mathbb{C}^\times$ as the unit circle.

$(3)$ $\mathbb{Q}^\times$, $\mathbb{R}^\times$ and actually $K^\times$ for any intermediate field extension $\mathbb{Q}\subseteq K\subseteq \mathbb{C}$.

$(4)$ $\mathbb{R}_{>0}$, $\mathbb{Q}_{>0}$, $K_{>0}$ for $\mathbb{Q}\subseteq K\subseteq \mathbb{R}$.

$(5)$ $S^1\times K_{>0}$ embedded as $\{Re^{i\theta}: R\in K_{>0}, \theta \in [0,2\pi]\}$, whenever $K_{>0}$ is as in $(4)$. This one might deserve some justification:

If $x_0=R_0 e^{i\theta_0}$ and $x_1=R_1e^{i\theta_1}$ are given for $R_0,R_1\in K_{>0}$ and $\theta_0,\theta_1\in [0,2\pi]$ then $x_0x_1=R_0R_1e^{i(\theta_0+\theta_1)}$ which is also in $S^1\times K_{>0}$. Inverses follow similarly, by $R_0\in K_{>0}$ implying $R_0^{-1}\in K_{>0}$. Of course, $1$ is in the set.

(6) $\mathbb{Z}/n\mathbb{Z}\times K_{>0}$ for $K_{>0}$ as in $(4)$ and $\mathbb{Z}/n\mathbb{Z}\hookrightarrow \mathbb{C}^\times$ as the $n^{th}$ roots of unity.

Is this list exhaustive? (I doubt it.)

I haven't thought too much about this, because I'm busy with other math things. However, if the answer to this is known I would appreciate a reference (an explanation would be great too, if this is actually easy).

  • $\begingroup$ Some thought: We can identify $\mathbb{C}^\star$ with $S^1\times\mathbb{R}^\star$ (using polar coordinates $z=re^{i\theta}$) so every subgroup of the first is a subgroup of the second. This gives you many subgroups of the form $G\times H$ where $G\leq S^1$ and $H\leq \mathbb{R}^\star$. $\endgroup$ – Yanko Jun 28 at 16:24
  • 2
    $\begingroup$ Should the identification not be $\mathbb{C}^\times \cong S^1\times \mathbb{R}_{>0}$? –– But otherwise, I agree that this is a good method of constructing subgroups. $\endgroup$ – Antonios-Alexandros Robotis Jun 28 at 16:26

Your classification is missing spirals: images of one-parameter homomorphisms $\exp(zt)$ when the complex number $z$ is not purely real or purely imaginary. Or discrete spirals, such as the image of the integers under such a homomorphism, or "dusty" spirals as the image of subgroups of $\Bbb Q$ under the homomorphism.

The subgroups of $\Bbb Q$ can be classified by "supernatural" rational numbers $n=\prod p^v$, which are formal infinite products of prime powers with exponents $v$ in $\Bbb Z\cup\{+\infty\}$. The subgroup corresponding to $n$ would be the set of all rationals with denominator dividing $n$ (with divisibility defined prime by prime, and every $p^v$ dividing $p^{\infty}$).

Note that $S^1\cong\Bbb R/\Bbb Z\cong(\Bbb Q^{\frak \oplus c})/\Bbb Z\cong(\Bbb Q/\Bbb Z)\times\Bbb Q^{\oplus\frak c}$, but picking a $\Bbb Q$-basis for $\Bbb R$ seems pretty wild - the axiom of choice guarantees the existence of a basis, but I don't know if anybody's ever constructed an explicit one. (Although apparently the "complexity" of such a basis is not as extreme as it could be made out to be, whatever that means.) And $\Bbb R_{>0}\cong\Bbb R\cong\Bbb Q^{\oplus\frak c}$ too, so the group $\Bbb C^{\times}$ is abstractly isomorphic to $(\Bbb Q/\Bbb Z)\times \Bbb Q^{\oplus\frak c}$.

Now any subgroup of this direct product will pull back (wrt the projection $\Bbb Q^{\oplus\frak c}\to(\Bbb Q/\Bbb Z)\times\Bbb Q^{\oplus\frak c}$) to a $\Bbb Z$-submodules (i.e. subgroups) of $\Bbb Q^{\oplus\frak c}$ containing $1$ in the "first" factor.

Probably we can go further with this by incorporating a version of Goursat's lemma, which would predict the spiral subgroups and likely classify all subgroups as products of spirals (where you count the circle and rays, as well as their discrete or "dust" versions, as generalized spirals).

  • $\begingroup$ @lisyarus Sorry, edit preview wasn't working on my deleted answer (which I deleted momentarily since it missed subgroups of $\Bbb Q$) so I used an "another answer" preview and then forgot to transfer back, so I lost your comment. I have seen "wild" used the way you mentioned in representation theory contexts, but I think (?) I have also seen wild used for noncomputable or undecidable problems in number theory contexts, and other contexts as just a normal English word. So I'll keep the word "wild" but avoid the technical term "wild problem." $\endgroup$ – runway44 Jun 28 at 17:04

More groups can be identified by taking a subring of $A\subset\mathbf{C}$ and then the group $A^*$ of invertible elements of that ring. For example $A$ could be the ring of integers of some algebraic number field, and Dirichlet's theorem gives their structure. Possibly $A$ can be a non-Noetherian ring.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.