Prove that $G$ has exactly $3$ subgroups iff $G$ is cyclic with $|G|$ = $p^2$
Here, $p$ is prime. Let us focus on the forward implication only.
My attempt:
Suppose $G$ has exactly $3$ subgroups. Then by definition there exists only one non-trivial proper subgroup of $G$. Let $H \subset G$ and take $g \in G$ s.t. $g \notin H$. Then by definition $<g> = \{e\}, G, \hspace{1mm}$ so $G$ is cyclic. By definition $|G|=n$ and since $G$ has $3$ subgroups, n = $pp'$ where $p,p'$ are prime. Suppose $p \neq p'$. Then, since a subgroup of a cyclic group is also cyclic, we can find $g_1, g_2 \in G$ s.t. $|<g_1>| = p$ and $|<g_2>| = p'$. But this is a contradiction since $G$ only has exactly $3$ subgroups. Thus $p=p'$ and $|G| = p^2$
I am not sure if I concluded correctly about letting $n=pp'$. Any feedback and critique appreciated. Thanks.
Best Answer
You seem to have some of the right ideas, but some steps in the proof are unclear. For instance, it's unclear how you conclude that $|G| = p^2$ or $n=pp'$.
Suppose $G$ has exactly $3$ subgroups. Then certainly $G \neq \{e\}$. Let $g \in G \setminus \{e\}$ and consider $\langle g \rangle$, noting that $|\langle g \rangle| \geq 2$. If $\langle g \rangle = G$, then $G$ is cyclic, but cyclic groups satisfy the converse of Lagrange, so $|G|$ has exactly three divisors, forcing $|G|=p^2$. On the other hand, if $\langle g \rangle \neq G$, then $\langle g \rangle$ is the unique proper, non-trivial subgroup of $G$. Uniqueness implies $|\langle g \rangle| = p$ for some prime $p$ and since $\langle g \rangle \neq G$, we get that $|G| = ap$ for some integer $a \geq 2$. Hence there exists some $g' \in G \setminus \{e\}$ such that $g' \notin \langle g \rangle$. In particular this means that $\langle g' \rangle \neq \{e\}$ and $\langle g' \rangle \neq \langle g \rangle$ but since $G$ has exactly three subgroups this forces $\langle g' \rangle = G$. Hence $G$ is cyclic, generated by $g'$, and by a similar argument as above we see that since $G$ is cyclic, it satifies the converge of Lagrange, and so it must have exactly three divisors, forcing $|G| = p^2$.