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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

    Hurkyl

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    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.
     
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