Finite groups with exactly one maximal subgroup

Solution 1:

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:

Let $G$ be a finite group. Suppose it has exactly one maximal subgroup. Then $G$ is cyclic.

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.

Solution 2:

If $G$ has only one conjugacy class of maximal subgroup, $M$, and is not itself a $p$-group, then for every prime $p$, $M$ contains a Sylow $p$-subgroup $P$ of $G$. But then by Lagrange, $|M|$ has order divisible by every prime power dividing $|G|$, so $|M|$ is a multiple of $|G|$. Since $|M| \leq |G|$, one must have $|M| = |G|$ and $M = G$, a contradiction. If $G$ is a $p$-group and one conjugacy class of maximal subgroup, then every maximal subgroup is normal, and so $G$ must have exactly one maximal subgroup, and so is cyclic as before.

In particular, every finite group has at least two classes of maximal subgroups. The non-abelian group of order 6 has exactly two classes.

It is not too uncommon for a finite group to have only one class of maximal subgroup that is not normal, and one that is normal (so two classes total): every finite primitive solvable group is like this. Depending on your definition, it might just be by definition. But there are a few definitions that make it sound exciting.

Solution 3:

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"?