1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 1 maximal subgroup -> prime order

  1. Dec 6, 2009 #1
    'Prove that if a finite group G has only one maximal subgroup M, then |G| is the power of a prime'

    I've somehow deduced that no finite group has only one maximal subgroup, and I'm having trouble seeing where I went wrong.
    This is what I have:

    Let [tex]H_1[/tex] be a subgroup of G. Either [tex]H_1[/tex] is maximal and equal to M or it is not maximal and there is a [tex]H_2[/tex] such that [tex]H_1<H_2<G[/tex] (using < to mean proper subgroup). Apply the same argument to [tex]H_2[/tex] and we get an ascending chain of subgroups. Since G is finite the process must end eventually (when we reach M).

    Thus every subgroup of G is a subgroup of M and hence every element of G is an element of M, and G=M, a contradiction.

    So would someone like to point out the flaw in the above reasoning? thanks.
    Last edited: Dec 6, 2009
  2. jcsd
  3. Dec 6, 2009 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Try substituting for G a particular group where the conclusion is obviously wrong. Do you know any groups with only one maximal subgroup?

    The most obvious ones to me to try are the two-element group, or maybe the cyclic group on 4 elements.
  4. Dec 6, 2009 #3
    Thanks Hurkyl. That should've been the first thing to do. For anyone interested, the problem is with the last sentence. It does not follow that every element of G is an element of M: the elements of G that generate G do not have to be in M.
  5. Sep 10, 2011 #4
    I know that this post is old, but since it is top result when i google this problem, I think it would be good if i give further detail of solution.

    Let $a \in G \setminus M$. Then $a$ generate $G$ which mean $G$ is cyclic. Suppose $|G| = p_{1}^{r_{1}}...p_{n}^{r_{n}}$ where $p_{i}$ is prime. So $a^{p_{i}}$ and $a^{p_{j}}$ generated a different maximal subgroup which is contradiction.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook