I was recently reading a proof in which the following property is used (and left as an exercise that I could not prove so far). Here is exactly how it is stated.
Let $G$ be a finite group. Suppose it has exactly one maximal subgroup. Then $G$ is cyclic.
Now, does this mean exactly one conjugacy class of maximal subgroups? If it really means exactly one maximal subgroup, then what can we say about a finite group that possesses exactly one class of (at least two) maximal subgroups that are all conjugate to each other?
Best Answer
Definition: A proper subgroup $M$ of a group $G$ is maximal iff whenever $M \leq H \leq G$, we have $H = M$ or $H = G$.
Alternative Definition: A subgroup $M$ of a group $G$ is maximal iff $M \neq G$ and there is no subgroup $H$ of $G$ such that $M \subset H \subset G$.
Proposition:
Proof:
Suppose $M < G$ is the maximal subgroup of finite group $G$. Take $a \in G\setminus M$.
If $H = \langle a\rangle < G$, then there is a maximal subgroup $K$ such that $H\subset K$. Clearly $K \neq M$, which leads to a contradiction to the uniqueness of the maximal subgroup. So $G = H$ must be cyclic. Furthermore, $G$ must be a $p$-group, since otherwise, $G$ would have more than one maximal subgroup.