TITLE: Finite groups with exactly one maximal subgroup
QUESTION [25 upvotes]: 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?

REPLY [4 votes]: The argument used to prove that a finite group with a unique maximal subgroup is cyclic can also be used to describe the structure of commutative local rings. I include this as an answer because it highlights a basic trick in algebra.
A ring $A$ is a (commutative) local ring if it has a unique maximal ideal $m$. Let $x\in A$, $x\not\in m$. We assert that $x$ is a unit in $A$. If not, the ideal $(x)=\{xa|a\in A\}$ generated by $x$ is a proper ideal of $A$ and is therefore contained in some maximal ideal of $A$. However, $m$ is the only maximal ideal of $A$ and this proves that $x\in (x)\subseteq m$; a contradiction. 
Exercise 1: Prove that if the set of non-units in a commutative ring $A$ is an ideal in $A$, then $A$ is a local ring.
Exercise 2: Let $x\in \mathbb{R}$ and consider the set of all ordered pairs $(f,U)$ where $U$ is open and $f:U\to \mathbb{R}$ is continuous. We define an equivalence relation on this set by setting $(f,U)\equiv (g,V)$ if $f|W=g|W$ for some open subset $W\subseteq U\cap V$. Let us define a ring structure on the set of all equivalence classes $A$ by the rules: $(f,U)+(g,V)=(f+g,U\cap V)$ and $(f,U)\times (g,V)=(f\times g,U\cap V)$. Prove that these operations are well-defined and prove that the ring $A$ in question is a local ring. (Hint: The set of all equivalence classes corresponding to representatives of the form $(f,U)$ such that $f(x)=0$ is the unique maximal ideal of $A$.) 
Exercise 3: Is the result of the above exercise true if the word "continuous" is deleted throughout? Is the result of the above exercise true if the words "continuous" and "open" are deleted throughout? Is the result of the above exercise true if the word "continuous" is replaced by either "differentiable", "smooth" or "analytic"?