Let G be a group with exactly two nontrivial proper subgroups. Show that G is cyclic. What are the possible orders of G?
Proof:
Assume G has no nontrivial subgroups.
Because G is trivial by the assumption above, then G is cyclic.
If G has exactly one nontrivial subgroups H, consider the subgroup generated by a nonidentity element g in G/H
Now suppose that H and K are the only nontrivial subgroups of G.
Recall that a group is never the union of two proper subgroups….
(Above is the start I have to my proof. I haven;t finished proving it is a cyclic group and can't figure out how to find what the possible orders may be.)
Best Answer
Let $H, K$ be the two non-trivial subgroups of $G$.
As you said, we cannot have $H \cup K =G$. Thus, there must exist some $x \in G \backslash (H \cup K)$.
As $x \neq e$ the only possibilities for $<x>$ are $H,K$ or $G$. Since the first two are not possible, we must have $<x>=G$.
To complete the proof, note that for a cyclic group of order $n$, there exists a subgroup of order $d$ for each $d|n$. So the question asks you to figure all the $n$ which have exactly $2$ non-trivial divisors.