Cyclic groups and subgroups Question from my Algebraic Structures Textbook: Suppose that $G$ is a group in which $\{1\}$ and $G$ are the only subgroups. Show that $G$ is finite and, in fact, is cyclic of order $1$ or a prime.
I am confused why the group would necessarily be finite. 
 A: This is a manifestation of the more general fact that any group with finitely many subgroups is in fact finite. See if you can justify this to yourself as follows: consider the cyclic subgroups generated by each element of the group $G$. There are only finitely many subgroups of $G$, so there are finitely many distinct cyclic subgroups. If any element of $G$ has infinite order, it generates a cyclic subgroup isomorphic to $\mathbb{Z}$, and therefore has infinitely many distinct subgroups, a contradiction. Hence, every element of $G$ has finite order, and $G$ has finitely many distinct cyclic subgroups. But $G$ is the union of the cyclic subgroups generated by its elements, so $G$ must be finite. 
In your particular example, one can cut this argument a bit shorter. If the only subgroups of $G$ are $G$ and $\{e\}$, then suppose $G$ is nontrivial, and pick a nontrivial element $g \in G$. Since the only subgroups of $G$ are $\{e\}$ and $G$, the cyclic subgroup $\langle g \rangle$ generated by $g$ must be all of $G$, whence $G$ is cyclic. If $g$ has infinite order, i.e. $G$ is infinite, then $G$ has a distinct cyclic subgroup for each positive integer $n$, namely the subgroup of $G$ generated by $g^{n}$, a contradiction, so $|G| < \infty$. The subgroups of any finite cyclic group are in bijective correspondence with the divisors of its order, so if $G$ has only trivial subgroups, then $|G|$ must be prime.  
