# Are proper subgroups of $\Bbb C^*$ cyclic?

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.

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

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 $H$.

• $\mathbb{Q}^*$ under multiplication is perhaps a simpler example of a countable subgroup that is not cyclic. – 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.

• Very interesting facts and references! – 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 $\{2^n : n \in \Bbb Z\}$.

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