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Alternating groups!

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data

    I was curious to know, say we have two even permutations taken out of A_4, say
    (12)(34) and (123), and we want to find the smallest subgroup of A_4 that contains both these permutations, then how would we go about it.
    This subgroup in this case will defenitely be A_4 itself, here is how i came to this conclusion.
    SInce that subgroup should contain both these permutations, then it also should contain the subgroups generated by those permutations, and also the elements that are derived when we multiply these by each other, i kept going this way, and i finally generated the whole A_4. BUt imagine if we were working with a group of higher order, since ordA_4 =12, then this would be a pain.

    SO my real question is this, is there any clever way of finding these subgroups that contain, like in this case, two other elements.

    If it were just for one, i know that the smallest subgroup would be the cyclic subgroup generated by that element, but what about this case???

    Any input is greately appreciated.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 13, 2008 #2
    So, does anyone knwo a clever way of doing it, or the way i did it is the only one?
  4. Oct 13, 2008 #3


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    There is no general purpose algorithm for doing such computations with groups. Heck, there are even groups for which we cannot generally compute (even in theory!) whether or not two elements are equal.

    Various techniques are useful in special cases. For example, if your group is a vector space, you have all the techniques of linear algebra at your disposal. Those same techniques even apply to computations with abelian groups, although there are some extra subtleties involved.
  5. Oct 13, 2008 #4

    So, you are saying that if i were asked to find the smallest subgroup, say now in A_8, that contains
    (12)(345) and (123)
    i have to explicitly compute all of them, untill i reach somewhere where the elements start to repeat themselves?

    This looks almost impossible to me, since the odds that we migt forget or do a mistake on the way are quite high, as far as i am concerned!
  6. Oct 13, 2008 #5
    Ok, for A_4 in particular i think i figured out a shortcut. by lagrange's theorem, we know that in A_4 there can be only subgroups of order 1, 2, 3, 4 or 6. since these are the only divisors of 12 which is the order of A_4. So, since it is required that both (12)(34) and (123) be in that subgroup we know that, that subgroup call it H, should contain also

    But clearly since this is not a subgroup, and it has defenitely more than 6 elements, which would be the subgroup with the highest order possible in A_4, besides A_4 itself, it means that A_4 will be the only such subgroup that will contain (12)(34) and (123).

    But again, this would be a nice method only if we are working with groups of not high order.
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