Unique Subgroup 5A8C8F

1. **Problem Statement:** Suppose $G$ is a finite group with exactly one nontrivial proper subgroup. We want to prove that $G$ is cyclic and that its order $|G| = p^2$ for some prime $p$. 2. **Key Idea and Formula:** Recall that a subgroup is a subset of $G$ that is itself a group under the operation of $G$. 3. **Step 1: Analyze the subgroup structure** Since $G$ has exactly one nontrivial proper subgroup, denote this subgroup by $H$. Then the subgroup lattice of $G$ is: $$\{e\} \subset H \subset G$$ where $e$ is the identity element. 4. **Step 2: $H$ is normal and unique** Because $H$ is the only nontrivial proper subgroup, it must be normal in $G$ (any conjugate of $H$ is also a subgroup of the same order, so it must be $H$ itself). 5. **Step 3: Order considerations** Let $|G| = n$ and $|H| = m$. By Lagrange's theorem, $m$ divides $n$. 6. **Step 4: $H$ is the unique subgroup of order $m$** Since $H$ is the only nontrivial proper subgroup, there are no other subgroups of any order dividing $n$ except $\{e\}$, $H$, and $G$. 7. **Step 5: $G$ is a $p$-group** If $n$ had more than one prime factor, then by Cauchy's theorem, there would be subgroups of prime order different from $H$, contradicting uniqueness. Thus, $n = p^k$ for some prime $p$ and integer $k \geq 1$. 8. **Step 6: $k=2$** If $k=1$, then $G$ is cyclic of prime order $p$ and has no nontrivial proper subgroups, contradicting the existence of $H$. If $k > 2$, then $G$ has subgroups of order $p$ and $p^2$ (by Sylow theorems), so there would be more than one nontrivial proper subgroup. Therefore, $k=2$. 9. **Step 7: $G$ is cyclic of order $p^2$** Groups of order $p^2$ are either cyclic or isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$. The latter has multiple subgroups of order $p$, contradicting uniqueness. Hence, $G$ is cyclic of order $p^2$. **Final conclusion:** $$G \cong \mathbb{Z}_{p^2}$$ where $p$ is prime. Thus, $G$ is cyclic and $|G| = p^2$.