# Cardinality Relation between a group and its subgroups

For a group $G$, let $F(G)$ denote the collection of all subgroups of $G$. Which one of the following situations can occur?

a) $G$ is finite but $F(G)$ is infinite

b) $G$ is infinite but $F(G)$ is finite

c) $G$ is countable but $F(G)$ is uncountable

d) $G$ is uncountable but $F(G)$ is countable

a) $G$ is finite means $\vert G \vert =n$, a finite number. so it has finite number of subgroups. so a) is false

c) can occur as this post shows.

My try for d):

take $G=C[0,1]$, the ring of all real valued continuous functions on $[0,1]$. It is a ring and in particular it is an uncountable additive group. It has uncountable number of maximal ideals, namely of the form $$H_\gamma=\{f \in C[0,1] : f(\gamma)=0\}$$ where $\gamma$ is any number in $[0,1]$.

These Uncountable $H_\gamma$'s are in particular additive subgroups of $C[0,1]$. So d) is false

My question is :

1) How to disprove b) ?

that is., How to prove an infinite group has infinite number of subgroups ?

2) Is my counterexample valid?

• Conisder the groups you can generate in the case of (b) and (d). An example where it is false does not show that it cannot occur just that it does not always occur Apr 13, 2018 at 15:27

To show (b) is false, let $G$ be a countable group. If $G$ contains an element $g$ of infinite order, then $\langle g^n\rangle$, $n\geq 1$ gives an infinite number of subgroups.
Therefore, assume that every element of $G$ has finite order. Then, the set $F_{\mathrm{fin}}(G)$ of finite subgroups of $G$ is already infinite. Indeed, $\{\langle g\rangle\mid g\in G\}$ must be an infinite set. If it were finite, then $G=\bigcup_{g\in G}\langle g\rangle$ would be finite being the finite union of finite sets.
For d), you've constructed one uncountable group with uncountable number of subgroups, but you have to prove it for arbitrary uncountable group, so I wouldn't call it a "counterexample". But considerations similar to those in linked question show that if you have a group of (infinite) cardinality $\kappa$, then it has at least $\kappa$ subgroups.