Prove that a group with exactly two proper nontrivial subgroups is isomorphic to $\mathbb{Z}_{pq}$ or $\mathbb{Z}_{p^3}$.

Solution 1:

Hint: If there are only two subgroups $H,K \subset G$ then $H \cup K \neq G$ (why?). Now what can you say about an element $g \in G \setminus (H \cup K)$?

Solution 2:

Suppose that $G$ has exactly two nontrivial proper subgroups $H$ and $K$.

If $K$ has a proper nontrivial subgroup, then $H \leq K$. Thus $K$ has exactly one proper nontrivial subgroup, which means that $K$ is cyclic of order $p^2$. Hence $G$ must be a $p$-group of order $p^3$, and any element $x \not\in K$ will generate $G$.

If $H$ and $K$ do not have proper nontrivial subgroups, $H$ and $K$ will be cyclic of prime order. It is clear that $H$ and $K$ will be normal subgroups and that $G$ is cyclic of order $pq$.

You try can classify groups with exactly $n$ subgroups for small $n$. This is the case $n = 4$. You can try to do the case $n = 5$ by using the classification of groups with exactly $n = 1, 2, 3, 4$ subgroups. A starting point is to notice that if a group with $n$ subgroups contains a subgroup with $n-1$ subgroups, it will be cyclic.

After that you can think about how the number of subgroups affects the structure of groups. What is the largest $k$ such that every group with $\leq k$ subgroups is cyclic? Abelian?

Solution 3:

Case 1: $|G|$ is not a $p-$group

Proof: Let $p,q$ be two distinct divisors of $|G|$. Let $N_1,N_2$ be sylow p,q-subgroups respectively. Clearly, these two subgroups are distinct (because their orders are different).

Claim: $|N_1|=p$

Proof: Suppose not. $Z(N_1)$ can't be trivial, because $N_1$ is a p-subgroup). If $Z(N_1)\subseteq N_1$, then we get a contradiction because $N_1,N_2,Z(N_1)$ will be three distinct proper non-trivial subgroups. If $Z(N_1)|=N_1$, then $N_1$ is abelian, thus $N_1^p,N_1,N_2$ are distinct non-trivial proper subgroups.

Similarly, $|N_2|=q$. Since there are two proper non trivial subgroups, thus all conjugates of $N_1$ are equal to $N_1$. Hence, $N_1\lhd G$. Thus, $N_1N_2$ is another subgroup of $G$. Thus, $N_1N_2$ must be equal to $G$. Since $N_1\cap N_2=\{e\}$. Hence $|G|=|N_1N_2|=pq$

Case 2: $|G|=p^n$

Proof: By Sylow theorems, we know that $G$ has non trivial proper subgroups of order $p,p^2,...,p^{n-1}$. Thus, $n\leq 3$. $n\not=2$ ,because $G$ would either be $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_p\oplus \mathbb{Z}_p$. $\mathbb{Z}_{p^2}$ has exactly one proper non-trivial subgroup. $\mathbb{Z_p}\oplus\mathbb{Z}_p$ has more than two non-trivial proper subgroups: $\langle(1,0)\rangle,\langle(1,1)\rangle,\langle(0,1)\rangle$. $n=0,1$ also lead to contradiction easily.