1. Not finding help here? Sign up for a free 30min 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!

Maximal subgroups of solvable groups have prime power index

  1. Apr 9, 2012 #1
    I would like to ask if somebody can verify the solution I wrote up to an exercise in my book. It's kind of long, but I have no one else to check it for me :)

    1. The problem statement, all variables and given/known data
    If [itex]H[/itex] is a maximal proper subgroup of a finite solvable group [itex]G[/itex], then [itex][G:H][/itex] is a prime power.


    2. Relevant equations
    Lemma 7.13 that I refer to is basicly this:
    http://crazyproject.wordpress.com/2...-finite-solvable-group-is-elementary-abelian/

    3. The attempt at a solution


    Statement is true for abelian groups, so only nonabelian solvable groups are considered in the proof. All finite nonabelian solvable groups have at least one normal group (the commutator) and therefore contain a minimal normal subgroup. By Lemma 7.13 (iii), these subgroups have prime power order.
    Assume there exists minimal normal subgroup [itex]N[/itex] such that [itex]N \not\subseteq H[/itex]. Since [itex]NH[/itex] is a subgroup of [itex]G[/itex] properly containing [itex]H[/itex], [itex]NH = G[/itex]. By Second Isomorphism Theorem, [itex]G/N = HN/N \cong H/(N \cap H)[/itex] and as all cardinalities are finite, [itex]\frac{|G|}{|N|} = \frac{|H|}{|N \cap H|}[/itex] which implies [itex]\frac{|G|}{|H|} = [G:H] = \frac{|N|}{|N \cap H|}[/itex], where the last one is a prime power.
    Assume all minimal normal subgroups of [itex]G[/itex] are contained in [itex]H[/itex] and let [itex]N[/itex] be one such. We work by induction. If [itex]|G| = 2[/itex], then it is of prime order and the maximal proper subgroup [itex]\langle e \rangle[/itex] has index 2, certainly a prime power. In the quotient [itex]G/N[/itex], [itex]H/N[/itex] is maximal since if [itex]H/N[/itex] is properly contained in some subgroup [itex]T/N[/itex] then it follows that [itex]H \subseteq T[/itex], so [itex]T = G[/itex]. Since order [itex]|G/N|[/itex] is strictly less than [itex]|G|[/itex], [itex][G/N:H/N] = \frac{[G/N]}{[H/N]} = \frac{[G:N]}{[H:N]} = p^n[/itex] by induction. Also the identity [itex][G:H][H:N] = [G:N][/itex] implies [itex][G:H] = \frac{[G:N]}{[H:N]}[/itex] so [itex][G:H] = p^n[/itex].
     
  2. jcsd
  3. Apr 9, 2012 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Seems ok.

    I also note that your proof also works in the abelian case, so there is no need to only look at nonabelian groups.
     
  4. Apr 10, 2012 #3
    Thanks for reading it and for the tip. An abelian group certainly has a normal minimal subgroup as long as it is not simple, and that is probably the case I could have handled separately.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Maximal subgroups of solvable groups have prime power index
Loading...