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Estimate number of groups

  1. Jun 6, 2012 #1
    how would you estimate number of groups of order 24? i do not need exact number, it is 15.

    I know that there are 5 groups of order 8 and there is 1 , 4 or maybe 7 sylow 3-groups. but i do not know what to do next
  2. jcsd
  3. Jun 6, 2012 #2
    There are some easy non-Abelian groups of order 24. We know that S3 has order 6, so

    S3 [itex]\bigoplus[/itex] Z4


    S3 [itex]\bigoplus[/itex] Z2 [itex]\bigoplus[/itex] Z2

    are non-Abelian groups of order 24. And of course so is S4.

    You can do the same trick with the non-Abelian groups of order 8, summed with Z3
  4. Jun 6, 2012 #3
    thanks, but i need some upper estimation of number of non-isomorphic groups of order 24 with some explanation.
  5. Jun 6, 2012 #4
    There are 15 groups of order 24, up to isomorphism. I'm not aware of the precise classification; I just have the number in some old notes.
  6. Jun 6, 2012 #5
    "Proof by old notes!" One of the most effective proof techniques. Right up there with proof by vague recollection; proof by wild handwaving; proof by "It's obvious!" and proof by unsupported claim :-)

    It's easy to get an upper bound. The group operation is a function from G x G to G; that is, it's a subset of G x G x G. So there can be at most 2243 possible group operations. That's a pretty weak upper bound, but it is an upper bound.

    Depending on the level that the question's being asked, that might be the answer. Perhaps they just want to see if you can come up with SOME upper bound that you can justify. And one way to attack any problem is to find SOME solution first, even a bad one; then try to find a better solution.
    Last edited: Jun 6, 2012
  7. Jun 6, 2012 #6
    Keep that up and I won't share my elementary proof of the Goldbach conjecture.
  8. Jun 6, 2012 #7
    proof by intimidation, proof by popular vote, proof by zero attendance, proof by impossible to follow argument (similar to wild handwaving), anyone, anyone, bueller?
  9. Jun 6, 2012 #8
    That just cost you an elementary proof of the Reimann Hypothesis (no complex analysis at all!)

    EDIT: Google reveals that Wiki happens to have a list...
  10. Jun 7, 2012 #9

    As there are [itex]\,p(3)=3\,[/itex] partitions of 3, and [itex]\,24=2^3\cdot 3\,[/itex] , there are 3 non-isomorphic groups of order [itex]\,24\,[/itex].

    As there are [itex]\,3\,[/itex] non-isomorphic non-abelian groups of order 8 there are at least 3 nonabelian groups of order 24, each containing one the above

    groups of order 8 as Sylow 2-subgroup.

    Now, I guess you can pick up some of them on the air: [itex]S_3\times C_4\,\,,\,S_3\times C_2\times C_2\,\,,\,A_4\times C_2\,\,,\,Q_8\times C_3\,\,,...[/itex] .

    Check all these are non-isomorphic.

    Of course, one can continue with semidirect products...for example, suppose there's a unique Sylow 2-subgroup [itex]\,P\,[/itex] , which is then normal, so

    that we can take one of the Sylow 3-sbgsp. [itex]\,Q\,[/itex] and make it act on [itex]\,\operatorname{Aut}(P)\,[/itex] by conjugation: [tex]q\cdot x=: x^q=q^{-1}xq\,\,,\,q\in Q\,\,,\,x\in P[/tex]

    It's not specially hard to see that this is a non-trivial action iff the group is non-abelian (which we can assume as the abelian

    groups are already sorted out) and this gives us a non-abelian group different from the ones listed above.

    In case [itex]\,P\,[/itex] is abelian we can also assume [itex]\,Q\,[/itex] is non normal.


    Ps The above is not, of course, an exhaustive listing neither of all the possible groups of order 24 nor of all the different constructions to ge them.
  11. Jun 8, 2012 #10
    Just to be clear (this was probably just a typo), it follows that there are 3 abelian groups of order 24.
  12. Jun 8, 2012 #11

    Of course, thanks...and it was the mother of all typos, in fact.

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