TITLE: If a group $G$ has only finitely many subgroups, then show that $G$ is a finite group.
QUESTION [11 upvotes]: If a group $G$ has only finitely many subgroups, then show that $G$ is a finite group.  I have no idea on how to start this question. Can anyone guide me?

REPLY [11 votes]: Suppose $G$ were infinite.  If $G$ contains an element of infinite order, then _.
Otherwise, every element of $G$ has finite order, and if $x_1, x_2, \ldots, x_n$ are any finite set of elements of $G$, then the subgroups $\langle x_i \rangle$ cover only finitely many elements of $G$.  Therefore __.

REPLY [5 votes]: I think the negative form of that is easier to prove: "If a group $G$ is infinite, then it has infinitely many subgroups".
What I would then do is take a member $a\in G$ such that $\left< a \right>$ is infinite, then prove there are an infinite number of subgroups just there, and say that if no such $a$ exists, there must be infinitely many subgroups of the type $\left< a \right>$.

REPLY [2 votes]: Hint: The infinite cyclic group $\mathbb Z$ has infinitely many subgroups ($n\mathbb Z$) so every element of $G$ must have finite order.  Use this to show that there is a finite collection $g_1, \ldots, g_n \in G$ of elements such that every element of $G$ is of the form $g_i^j$ for some choice of $i$ and $j$.