A group with finitely many subgroups is finite.

Let $G$ be a group with finitely many subgroups. For each $x\in G$ we define $G_x:=\langle x\rangle:=\{x^n:n\in\mathbb{Z}\}$. Is clear that $$G=\bigcup_{x\in G}G_x$$
Suppose that $|G_x|=\aleph_0$ for some $x\in G$, then we can conclude that $G_x\cong\mathbb{Z}$ and this implies that $G_x$ has infinitely many subgroups in contradiction with the initial hypotesis. Then for any $x\in G$, $G_x$ must be finite. Now, exists $\{x_1,x_2,\dots, x_n\}$ such that $$\bigcup_{x\in G} G_x=\bigcup_{1\leq k\leq n}G_{x_k}$$ Because $G$ has only finitely many subgroups, so $G$ is finite.

Is my proof right?

• You are correct that every $G_x$ must be finite, but that's not enough to conclude the whole group is finite. You also need to use that the $G_x$ cannot all be distinct groups. – Cheerful Parsnip May 25 '18 at 20:29
• @CheerfulParsnip See my edit please, I think that I fix my mistake. – Gödel May 25 '18 at 20:35
• @Shaun Yes, I know that there exists another sizes of infinite but note that each $G_x$ is index by $\mathbb{Z}$. – Gödel May 25 '18 at 20:37
• I would use that each element generates a cyclic subgroup (which contains the element), so the union of the cyclic subgroups is the whole group. An infinite cyclic group has infinitely many distinct subgroups, and these would be subgroups of $G$ - so all the cyclic subgroups are finite. (each element of an infinite cyclic group is the identity or can be paired with its inverse, and the distinct pairs generate distinct subgroups). Then the finite union of finite sets is finite. – Mark Bennet May 25 '18 at 20:57
• Gödel: looks good! Pretty close to what @Mark Bennet wrote in his comment. – Cheerful Parsnip May 25 '18 at 21:50

Your proof is essentially good, but it can be streamlined.

Let $x_1,\dots,x_n\in G$ such that $\langle x_1\rangle,\dots,\langle x_n\rangle$ are all the cyclic subgroups of $G$. Then, since every element of $G$ belongs to a cyclic subgroup, we have $$G=\bigcup_{i=1}^n\langle x_i\rangle$$ Moreover, every cyclic subgroup of $G$ is finite, because if $\langle x\rangle$ is infinite, it has infinitely many (cyclic) subgroups and each such subgroup is also a subgroup of $G$.

Therefore $G$ is finite.

I see nothing wrong in it, but it is incomplete. You did not justify that if $G_x\simeq\mathbb Z$, for some $x\in G$, then $G$ has infinitely many subgroups.

• Because each subgroup of $G_x$ is subgroup of $G$ too. For this case, as $\mathbb{Z}$ has infinitely many subgroups then $G_x$ too. – Gödel May 25 '18 at 20:40
• @Gödel I know that. I was just saying that this should be part of your proof. – José Carlos Santos May 25 '18 at 20:42
• Ok, thanks for your observation. – Gödel May 25 '18 at 20:43

Here's a shorter proof. If a group has finitely many subgroups it also has finitely many cyclic subgroups, so we have that $\{\langle x \rangle | x \in G \}$ is a finite set. Now if $G$ is infinite there by above assertation one $\langle x \rangle$ is infinite, as $G = \bigcup_{x \in G}\langle x \rangle$

Now it's not hard to see that $\langle x^m \rangle \not = \langle x^n \rangle$ for positive $m\not=n$ and this means we have infinitely many subgroups of $G$, which is wrong.

• The last part is not true if $n=-m\neq 0$ – Mark Bennet May 25 '18 at 21:13
• @MarkBennet Oh, yeah. Anyway you can consider both $m,n$ positive :) – Stefan4024 May 25 '18 at 21:15