Finite and any two maximal subgroups intersect trivially implies not simple non-abelian

Statement

Suppose $G$ is a finite group with the property that any two distinct maximal subgroups of $G$ intersect trivially. Then, $G$ is not a simple non-abelian group.

Related facts

Direct applications

• Finite non-abelian and every proper subgroup is abelian implies not simple
• Finite non-nilpotent and every proper subgroup is nilpotent implies not simple

Indirect applications

• Classification of cyclicity-forcing numbers
• Classification of abelianness-forcing numbers
• Schmidt-Iwasawa theorem: This states that if every proper subgroup of a finite group is nilpotent, the group itself is a finite solvable group

Facts used

1. The trivial subgroup is maximal if and only if the group is a group of prime order, which is a simple abelian group.
2. Lagrange's theorem
3. Group acts as automorphisms by conjugation: Thus, conjugates of a maximal subgroup are maximal and all have the same order.
4. Size of conjugacy class of subgroups equals index of normalizer
5. In a finite non-cyclic group, every element is contained in a maximal subgroup. This basically follows from the fact that cyclic iff not a union of proper subgroups.

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

We prove the statement by contradiction.

Given: A finite simple non-abelian group $G$ of order $n$ such that any two distinct maximal subgroups of $G$ intersect trivially.

To prove: A contradiction.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Every maximal subgroup of $G$ is self-normalizing	Fact (1)	$G$ is simple non-abelian		[SHOW MORE] First, note that $G$ is not simple abelian, so the trivial subgroup is not maximal in it. Thus, every maximal subgroup is nontrivial. Since $G$ is simple non-abelian, a maximal subgroup of $G$ cannot be normal, and hence, its normalizer must equal itself. Note that we are using the fact that the trivial subgroup of $G$ cannot be maximal since that would force $G$ to be a cyclic group of prime order, which is abelian.
2	If $A$ is a maximal subgroup of $G$ with $a$ elements, and $G$ has $n$ elements, the union of all conjugates of $A$ in $G$ has $n-(n/a)+1$ elements	Facts (2), (3), (4)	$G$ is finite, has order $n$ Any two maximal subgroups of $G$ intersect trivially	Step (1)	[SHOW MORE] By Step (1), $N_{G}(A)=A$, so the number of conjugates of $A$ in $G$ is $[G:N_{G}(A)]=n/a$. Each of these is maximal, and we are given that any two of them intersect in the identity element. Thus, each subgroup has $a-1$ elements not in any of the others. The total number of elements is $(n/a)(a-1)+1=n-(n/a)+1$.
3	The union of conjugates of any one maximal subgroup has at least $1+(n/2)$ of the elements, but does not have all the elements			Step (2)	[SHOW MORE] Note that $a$ divides $n$ and is not equal to $n$, so $n-(n/a)+1<n-(n/n)+1=n$. Also, $a\neq 1$, since if the subgroup of order $1$ is maximal, then that would mean that the whole group is cyclic of prime order and hence abelian. So, $a\geq 2$, hence the number of elements is at least $n-(n/2)+1=(n/2)+1$.
4	We have the desired contradiction, because cardinality considerations force $G$ to have more than one conjugacy class of maximal subgroups, but there is not enough room for two conjugacy classes of subgroups.	Fact (5)	Any two maximal subgroups of $G$ intersect trivially.	Step (3)	[SHOW MORE] Since the union of all conjugates of one maximal subgroup (say $A$) does not cover the whole group, there exist elements outside the union of the conjugates of $A$. Since $G$ is non-abelian and hence not cyclic, this element generates a proper subgroup of $G$ that lies in some maximal subgroup, say $B$. The union of the conjugates of $B$ has at least $(n/2)+1$ of the elements of $G$, and so does the union of the conjugates of $A$. But from the given data, no conjugate of $B$ can intersect any conjugate of $A$ at any non-identity element. Thus, these two sets of size at least $(n/2)+1$ intersect in a set of size $1$. By the principle of inclusion and exclusion, the union of these sets has size at least $2((n/2)+1)-1=n+1$, a contradiction since the size can be at most $n$.