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

So here it goes:


$G$ is a group

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


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.

  • 1
    $\begingroup$ Use $\langle X\rangle$ for $\langle X\rangle$. $\endgroup$
    – Shaun
    Sep 24, 2020 at 13:49
  • 1
    $\begingroup$ There's a proof here. $\endgroup$
    – Shaun
    Sep 24, 2020 at 13:54
  • 5
    $\begingroup$ @Everstudent Your argument rests on the false claim that the existence of infinitely many subgroups implies the existence of an element of infinite order. The product of infinitely many groups of order $2$ is a counterexample to that assertion. $\endgroup$ Sep 24, 2020 at 13:57
  • 2
    $\begingroup$ My argument uses the existences of a finitely many subgroups in an infinite group which is a contradiction. I suppose that G is infinite but has a finite number of subgroups. Then I look at all the cyclic subgroups of G, they have to be a finite number obviously. And their union has to be G. But then at least one of the cyclic groups has to be infinite. $\endgroup$ Sep 24, 2020 at 14:00
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    $\begingroup$ The proof is correct. $A_G$ is a union of finitely many elements. If all $A\in A_G$ were finite, then $G$ would be finite. Hence there must be one element $A=\langle x \rangle$ which is infinite. $\endgroup$ Sep 24, 2020 at 14:05

1 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)

  • 1
    $\begingroup$ I thought this might be of interest since the torsion of a group is a very important way to understand infinite groups. $\endgroup$
    – Alon Yariv
    Sep 24, 2020 at 14:22
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    $\begingroup$ The set of torsion elements is not necessarily a subgroup when the group is nonabelian. $\endgroup$
    – Jim
    Sep 24, 2020 at 15:53
  • $\begingroup$ Sorry, that is indeed true. I corrected the argument $\endgroup$
    – Alon Yariv
    Sep 24, 2020 at 16:39
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    $\begingroup$ The idea of cosets isn't entirely well defined here. Given a non-subgroup subset translations of that set don't form a partition of the group. This means, in particular, that the index is not necessarily well defined and your claim that it is "multiplicative" isn't necessarily a valid one. $\endgroup$
    – Jim
    Sep 24, 2020 at 16:56
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    $\begingroup$ It seems like this is all fixable, you just want to argue that either all elements have finite order, in which case you use your inductive argument, or some element has infinite order, in which case you use your $\mathbb Z$ argument. But I don't think it makes sense to talk about indexes here. $\endgroup$
    – Jim
    Sep 24, 2020 at 16:58

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