Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
It means "exactly one maximal subgroup". –  Arturo Magidin May 18 '11 at 19:49
2  
Though, as Jack points out, the weaker claim "one conjugacy class of maximal subgroups" in fact implies that there is only one maximal subgroup. –  Arturo Magidin May 18 '11 at 21:02
    
Let G be a finite group.prove that every proper subgroup of G is in a maximal subgroup –  user76834 May 8 '13 at 14:58

4 Answers 4

up vote 12 down vote accepted

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.

share|improve this answer
1  
Two more proofs (including the reduction from an infinite group to the finite case) are given at artofproblemsolving.com/Forum/viewtopic.php?f=61&t=404462 –  Jack Schmidt May 18 '11 at 20:44
1  
Another way of proving it is as a corollary to the fact that, if $M$ is a maximal subgroup of a finite group $G$, then $G$ cannot be the union of the conjugates of $M$. To prove that, consider the permutation representation of $G$ on the cosets of $M$ and use the fact that a nontrivial transitive permutation representation of $G$ must contain fixed-point-free permutations. –  Derek Holt May 19 '11 at 9:23
    
If I understand correctly, your second paragraph should refer to "every finite non-cyclic group". –  Zev Chonoles Jun 24 '11 at 20:52

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.

share|improve this answer
    
use \langle and \rangle for the angle brackets. –  Arturo Magidin May 18 '11 at 21:01
    
You'll want to bookmark DeTeXify (drawing an angle bracket brings \langle at the top or almost at the top for me), and/or pages with lists of LaTeX symbols, such as this one. –  Arturo Magidin May 18 '11 at 21:09
    
@Arturo: Yes, I believe I have it bookmarked...but I've been on a bookmarking frenzy...and need bookmarks to bookmark my bookmarks! I'll put it on my toolbar ;-) –  amWhy May 18 '11 at 21:33

The following fact implies this (it's already covered in the answers above, so this isn't really an original contribution):

http://groupprops.subwiki.org/wiki/Union_of_all_conjugates_is_proper

One application of this is that a finite non-abelian group in which every subgroup is abelian cannot be simple:

http://groupprops.subwiki.org/wiki/Finite_non-abelian_and_every_proper_subgroup_is_abelian_implies_not_simple

share|improve this answer
    
@ Vipul: you should develop more your answers, not simply give links to your website, although it is certainly well done. The object here is to provide self-contained answers (as far as this is reachable in space and time). –  Thomas Connor May 24 '11 at 6:49
    
@Thomas: I don't really see the need to provide an answer, if a link contains exactly the answer for the given question –  user9413 May 24 '11 at 18:16
    
I don't mean that the answer is not on his website. What I mean is that this is not a place to make advertisements for his website (what he does in every post he writes). Answers need to be self contained in this forum. –  Thomas Connor May 24 '11 at 19:16

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"?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.