
Describe all groups $\mathcal G $ whose lattices of subgroups look like the Hasse Diagram shown.
I do not understand how to begin and I am unable to visualize the lattice relation among the subgroups. Some help in that direction will be greatly appreciated.
According to the given lattice diagram of subgroups of some group $G$, $G$ has a unique nonidentity proper subgroup. It can be shown that such a group $G$ is necessarily a cyclic group of order the square of a prime. For suppose the order of $G$ is divisible by two distinct primes $p$ and $q$. Then, by Cauchy's theorem $G$ contains (cyclic) subgroups of order $p$ and $q$, which are necessarily distinct, but your lattice diagram shows only one nonidentity proper subgroup. Thus, the order of $G$ is $p^m$ for some prime $p$ and some $m \ge 2$ (here, $m \ne 1$ because the cyclic group $C_p$ has no nontrivial subgroups by Lagrange's theorem). Any group of order $p^m$ contains subgroups of orders $1,p,p^2,\ldots,p^{m-1}$ (this can be proved using the isomorphism theorems), whence $m=2$ by the given lattice diagram. Thus, $|G|=p^2$. A group of order $p^2$ is either the cyclic group $C_{p^2}$ or the direct product $C_p \times C_p$. But the latter group has more than one nonidentity proper subgroup, contradicting the lattice diagram. Thus, $G$ must be $C_{p^2}$.
A simpler proof is as follows. Suppose the group $G$ of the given lattice is not cyclic. Then, $\exists x \in G-1$ such that $\langle x \rangle$ is a proper subgroup of $G$. Let $y \in G - \langle x \rangle$. Then $\langle y \rangle$ is also a proper subgroup of $G$ and is distinct from $\langle x \rangle$, contradicting your lattice diagram. Thus, $G$ must be cyclic. Since $G$ has only one nontrivial subgroup, by the basic results on all subgroups of cyclic groups, the order of $G$ must have only one divisor besides 1 and $|G|$. Hence this order must be the square of a prime.