TITLE: A group $G$ has a finite number of subgroups if and only if $G$ is finite.
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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)