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Smallest non-solvable group

  1. Jan 1, 2004 #1

    as part of a referate i prepare, i have to do an induction proof about the group order of non solvable groups.
    however the publication i have to present proceeds this proof without giving a starting case and neither did i easily find one in my litarature.
    maybe this is a very trivial question, but im not that familiar with this and for me it isnt.
    so could anyone help me finding a start for my induction? maybe a dyhedral group?
    in addition i need to show that for this Group that G = G'(where G' is the Commutatorgroup) and also ({y}, G) = G for any y from G(in my paper this is called commutator simple, but i got no clue, if this is a common definition), the group should also not be too huge ;)
    any help would be appreciated.
  2. jcsd
  3. Jan 1, 2004 #2


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    What do you mean by 'non-solvable'?

    I expect that the group with one element meets all of the requirements that you listed, but is not suitable for your proof?
  4. Jan 1, 2004 #3
    ok. here some explanations.
    first of all i need to define what a commutator is.

    lets define [a, b] as a*b*a^-1*b^-1. let us call this the commutator from a and b.

    similary for 2sets A, B we call [A, B] = {a*b*a^-1*b^-1 | a from A, b from B} the commutator from A and B.

    so for a group G we can determine [G, G]. note that this isnt a group again in general.

    now we define G' = K(G) as the smallest subgroup from G which contains [G, G]. and call it the commutator subrgoup.

    for easier understanding one should also notice that for elements from the centre of G(which contains all the commutative elements...there for each element a from the centre a*b=b*a is true for all b from G) the commutator is always the idendity.

    therefore the commutator subgroup is a measurment of how "abelian" a group is(for a abelian group its just the id.

    Now we define K(2)(G) as K(K(G)) and K(n)(G) as K(K(n-1)(G)).

    since our group is of finite size we can now obviously derrive a series:
    G > K(G) > ... > K(n-1)(G) = K(n)(G).
    it should be easy to observe that (if at all) the size of commutator subgroup differs from the one of its origin it can only be smaller and on the other hand the id is always a member of the commutator subgroup.

    now we call a group G solvable group, if this series terminates in K(n)(G) = {id}.and ofcourse we call groups which dont satisfy this non-solvable groups. this series always terminates in maximum the group order steps.

    as an example that solvable groups arent identically to the abelian Groups you can look at S4.
    K(S4) = A4
    K(A4) = V4
    K(V4) = {id}
    therefore S4 is non-abelian, but

    now since obviously all abelian groups are solvable, since K(G) = {id}, if G is abelian.

    the group consisting only of one element is ofcourse solvable too. all Symmetric groups for n>=5 are non-solvable, but the order of S5 is allrdy 120.
    now since we have K(Sn) = An, An ofcourse is not solvable either, but still has order 60.
    the question is, if this one is allrdy the smallest.
    im actually hoping for another one, since i also need to prove that K(G) = [{y}, G] for each y from G. hereby [{y}, G] describes the smallest subgroup containing [{y}, G] ofcourse or look for another way to prove my base case. in anyway i need the smallest non-solvable group first ;)

    P.S. i hope its understandable, obviously english isnt my native language ;)
    Last edited: Jan 1, 2004
  5. Jan 10, 2004 #4


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    The term "solvable" is from Galois theory: a polynomial equation is "solvable by radicals" if and only if its Galois group is a solvable group.

    Since, for any positive integer n, there is an nth order polynomial equation such that its Galois group is Sn. Since Sn is NOT solvable for n>4, polynomials of degree 5 or greater are not all solvable by radicals (and so there can be no formula for their solution).

    S5 which has order 5!= 120 is NOT solvable- that's about the best I can do!
  6. Jan 10, 2004 #5
    most likely A5 is the smallest non-solvable group. its atleast non-solvable and K(A5)=A5 as well as commutator simple. it has order 60.
    i cant prove its the smallest, but i couldnt find a smaller one.
    for my referate though i found another solution and it went very well ;)
  7. Jan 12, 2004 #6


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  8. Jan 16, 2004 #7

    matt grime

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    here are some facts you might find useful

    let p,q,r be distinct primes

    every group of order p^a.q^b is solvable,
    every group of order pqr is solvable

    there is therfore no non-solvable group of order less than 60.

    prove there is a non-solvable group of order 60. hint, it suffices to show there is a simple non-abelian group of this order, which can be done simply by examining conjugacy classes (easy) or by other methonds (harder)
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