A group $G$ has a finite number of subgroups if and only if $G$ is finite.

Question

I found this as an exercise, and wrote my own solution but am interested in a shorter/easier one.

So here it goes:

Statement:

$G$ is a group

$G$ has a finite number of subgroups <=> $G$ is finite.

Proof:

Suppose $G$ has an infinite number of elements but a finite number of subgroups.

Let's look at the cyclic subgroups of $x$ where $x \in G$.

$A_G=\{\langle x\rangle : x \in G\}$

Since the elements of $A_g$ are subgroups of $G$ => $A_G$ has a finite number of elements.

Obviously $\cup_{A \in A_G}{A} = G$.(since every $x \in G$ would belong to $\langle x\rangle$ which is in $A_G$.

So it's a given that for some $x \in G$, $\langle x\rangle$ must have an infinite number of elements.

But then we can make infinitely many subgroups of $\langle x\rangle$ like: $\langle x^2\rangle$,$\langle x^3\rangle$,$\langle x^4\rangle$,etc.(which are all different, but to convince oneself, we can only look at $\langle x^p\rangle$ where p is prime.)

Hence G has an infinite amount of subgroups which is a contradiction, so G has to be finite.

Now in the opposite direction:

Suppose G is finite. Let $|G|=n$.

$P(G)$(the powerset of G) will have only $2^n$ elements. But the set of subgroups of G is a subset of $P(G)$.

Hence G has a finite number of subgroups.

Answer 1

The way I manage to prove it is as follows:

For a group $G$ we define the Torsion as the set of elements of finite order: $$\text{Tor}(G)=\{g\vert\exists n\in\mathbb{N}\;s.t.\;g^n=1\}$$ Now let us consider the two cases:

case 1: $G= \text{Tor}(G)$

In this case we have $\text{Tor}(G)$ being an infinite set. Since every element $t\in \text{Tor}(G)$ is of finite order we can inductively generate $\aleph_0$ finite cyclic subgroups: $$t_n\in \text{Tor}(G)-\cup_{i=1}^{n-1}\langle t_i\rangle \:\text{and take the subgroup} \: \langle t_n\rangle$$

case 1: $G\neq \text{Tor}(G)$ is infinite

This case implies the existence of an element $g\in G-\text{Tor}(G)$ that is not of finite order. In this case, $\langle g\rangle$ is a cyclic group of order $\aleph_0$ which is isomorphic to $\mathbb{Z}$, yielding an infinite amount of subgroups.

Let me just add that a finite group has a finite number of subgroups, for there are finitely many subsets (for completeness of argument)