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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.)

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    $\begingroup$ The first question was adequately discussed here. $\endgroup$ – Jyrki Lahtonen Jun 8 '16 at 16:21
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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.

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This is exercise 35 in the additional exercises to chapter 1-4 in Gallian's Contemporary Abstract Algebra.

The possibilities for $|G|$ are $pq$ and $p^3$ for some primes $p,q$.

To see this consider a group $G$ with exactly two non trivial proper subgroups and let $g\in G \setminus \{e\}$. Then $\langle g \rangle$ is a subgroup. If $\langle g \rangle = G$ then $G$ is cyclic.

If $\langle g \rangle \subsetneq G$ let $h \in G\setminus \langle g \rangle$. Then again $\langle h \rangle $ is a subgroup and if $\langle h \rangle = G$ then $G$ is cyclic. If $\langle h \rangle \subsetneq G$ note that since $\langle g \rangle \cup \langle h \rangle \subsetneq G$ there is $x \in G \setminus \langle g \rangle \cup \langle h \rangle$. Since $\langle x \rangle$ is a subgroup that is neither equal to $\langle g \rangle$ nor to $\langle h \rangle $ it must be true that $\langle x \rangle = G$ hence $G$ is cyclic.

Now we have shown that $G$ is cyclic. The claim about the order of $G$ follows directly from the fact that in a cyclic group there is a subgroup for each divisor of the order of the group.

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