Let $\Bbb C$ be the field of Complex numbers and $\Bbb C^*$ be the group of non zero Complex numbers under multiplication. Then every proper subgroup of $\Bbb C^*$ is cyclic. Is it correct statement?

I know that every finite subgroup of $\Bbb C^*$ is cyclic. But I am doubtful about above statement. Please clarify my doubt. If $\Bbb C^*$ has infinite proper subgroup then it will not be cyclic. But I don't have any example. Please correct me if I am wrong.

| cite | improve this question | | | | |

No. Let $\mathbb{R}^{\ast}$ be the subgroup of nonzero real number under multiplication. This subgroup is not cyclic because it is uncountable.

| cite | improve this answer | | | | |

Like pisco125's answer, the subgroup $\{z \in \mathbb{C} \ : \ |z| = 1 \}$ cannot be cyclic since it's uncountable. But there's even a countable subgroup of this subgroup that is not cyclic.

Consider $H = \{z \in \mathbb{C} \ : \ |z| = 1 \text{ and } z^n = 1 \text{ for some }n \in \mathbb{Z}^+\}$. One can verify this is indeed a subgroup: if $w_1^k = 1$ and $w_2^m = 1$, then $(w_1w_2)^{km} = 1$. Checking for inverses is similarly easy.

However, no $w \in H$ can generate. For any $w \in H$, we'll have $w^n = 1$ for some $n \implies w$ can only generate $n$ distinct elements of $H$.

| cite | improve this answer | | | | |
  • 3
    $\begingroup$ $\mathbb{Q}^*$ under multiplication is perhaps a simpler example of a countable subgroup that is not cyclic. $\endgroup$ – gj255 Jun 15 '17 at 13:31

Others have already given the easiest examples of noncyclic subgroups of $\mathbb{C}^{\times}$.

In fact, by "polar coordinates" we have

$\mathbb{C}^{\times} \cong \mathbb{R}^{>0} \times S^1$

Each of $\mathbb{R}^{> 0}$ and $S^1$ are uncountable, hence very far from cyclic.

The point of my answer is to record my first thought upon seeing the question:

Surely every commutative group in which all proper subgroups are cyclic must be countable.

A little checking confirms that this is true. In the paper

Xu, Maoqian(PRC-GDUT-TM) Groups whose proper subgroups are Baer groups. (English summary) A Chinese summary appears in Acta Math. Sinica 40 (1997), no. 1, 154. Acta Math. Sinica (N.S.) 12 (1996), no. 1, 10–17.

(it is freely available online, though I couldn't find a link that works directly)

I found

Lemma 2.1: A commutative group in which every pair of proper subgroups generates a proper subgroup is either a cyclic $p$-group or a quasi-cyclic group -- a.k.a. a Prüfer $p$-group.

Thus in an uncountable commutative group $G$, there is a proper uncountable subgroup -- otherwise every pair of proper subgroups would be countable, thus generate a countable, hence proper subgroup, and $G$ would be a cyclic $p$-group (finite) or a quasi-cyclic group (countably infinite).

However the proof given there refers to Robinson's GTM on group theory (okay) and a 1964 paper of Newman and Wiegold (less okay). I wonder if there is a self-contained, elementary derivation. Note though that it is also tempting to believe that any uncountable group has a proper uncountable subgroup. This was disproved by Shelah: see here.

| cite | improve this answer | | | | |
  • $\begingroup$ Very interesting facts and references! $\endgroup$ – pisco Jun 17 '17 at 13:05

No, not every proper subgroup of $\Bbb C$ is cyclic. For example, consider the complex numbers with rational real/complex parts, i.e. $G = \{x + iy : (x,y) \in \Bbb Q^2 \setminus \{(0,0)\}\}$.

In fact, $\Bbb C^*$ does have infinite cyclic subgroups. For instance, consider $\{2^n : n \in \Bbb Z\}$.

| cite | improve this answer | | | | |
  • $\begingroup$ Thanks everyone for your time. $\endgroup$ – ASHWINI SANKHE Jun 15 '17 at 6:23
  • $\begingroup$ Could you possibly mean $\{ 2^n \ : \ n\in \mathbb{Z}\}$ (you have no inverse elements otherwise)? $\endgroup$ – Severin Schraven Jun 15 '17 at 9:44

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.