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Find A Permutation of order 6 in S5

  1. Nov 5, 2003 #1

    First I will pose my question:

    I am not sure what this question is asking. Is this question asking for a subgroup of S5 with that consists of six elements?

    Any help is appreciated. Thankyou.
    Last edited by a moderator: Nov 5, 2003
  2. jcsd
  3. Nov 5, 2003 #2
    I am trying the best I can to do this question without really knowing what I am doing so forgive my fumbling around please and thankyou.

    Let A be a subgroup of S5. And let

    a1, a2, a3, a4, a5, u where u = a6 which is the identity element.

    be elements of A.

    Now I know that

    a1 o a5 = u
    a2 o a4 = u
    a3 o a3 = u.

    This confirms the existence of inverses - each element of A has an inverse.

    I also know that

    a1 o a1 = a2
    a2 o a2 = a4
    a3 o a3 = u
    a4 o a4 = u o a2 = a2
    a5 o a5 = u o a4 = a4

    I also know that for every element in A undergoing a binary operation yields an element of A.

    Not sure what to do next however.

    As my work progresses I will post more.

    Any input/insight is appreciated. Thankyou.
    Last edited by a moderator: Nov 5, 2003
  4. Nov 5, 2003 #3


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    Well, 6 = 2 * 3. So, if you could write down a permutation that has two commuting parts, one of order 2 and one of order 3, what's the order of their product?
  5. Nov 5, 2003 #4
    I would guess that IF I could find and write down a permutation with two commuting parts it's product would have an order of six(?)

    However, I don't know how to find a permutation that has two commuting parts, one of order 2 and one of order 3. And I don't know why if I found a permutation that has two commuting parts, one of order 2 and one of order 3, why it would have an order of six.

    Thanks for your help anyway Hurkyl. I appreciate it. But I don't think I have an aptitude for this course.

  6. Nov 5, 2003 #5


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    It's asking you to find an element π of S5 such that π6=identity but no lower power of π is the identity.

    Alternatively, this is equivalent finding a subgroup that is isomorphic to Z6
  7. Nov 5, 2003 #6


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    All right, I have two exercises for you, wubie.

    Exercise 1:

    Let x and y be commuting elements of a multiplicative group, and x has order 2 and y has order 3.

    Let z = xy and write the first 12 powers of z in the form:
    z^k = x^i * y^j

    And simplify as much as possible.

    Exercise 2:

    Determine a sufficient condition for two permutations to commute by randomly picking cycles and multiplying them. (You can probably figure it out after a few dozen examples)
  8. Nov 5, 2003 #7
    Alright. I have pretty much given up on this course (I am thinking about dropping it), but I will give the exercise a shot (since I do not like giving up). I will approach this exercise by giving some elementry definitions to see if I understand the question(s).

    Part 1

    Multiplicative Group:

    A non-empty set in which properties of a group apply and whose binary operation is multiplication.

    Commutative Property of the group:

    Given x, y which are elements of the Group G, x o y = y o x in which the product of x,y is an element of the group G.

    Order of an element:

    The order of an element a which is an element of G is the least positive integer n for which an = u (the identity element).

    So from what I know (which is not much) I think that if x is of order 2 then

    x2 = u


    y3 = u.

    Given that z = xy

    then the first 12 powers of z are:

    z1 = x y
    z2 = x2 y2 = u y2 = y2
    z3 = x3 y3 = x2 x u = u x u = x
    z4 = x4 y4 = (x2)2 y y3 = (u)2 y u = y
    z5 = x5 y5 = (x2)2 x y2 y3 = u x y2 u = x y2
    z6 = x6 y6 = (x2)3 (y3)2 = (u)3 (u)2 = u u = u
    z7 = x7 y7 = (x2)3 x (y3)2 y = (u)3 x (y)2 y = x y
    z8 = x8 y8 = (x2)4 (y3)2 y2 = (u)4 (u)2 y2 = u u y2 = y2
    z9 = x9 y9 = (x2)4 x (y3)3 = (u)4 x (u)3 = u x u = x
    z10 = x10 y10 = (x2)5 (y3)3 y = (u)5 (u)3 y = u u y = y
    z11 = x11 y11 = (x2)5 x (y3)3 y2 = (u)5 x (u)3 y2 = u x u y2 = x y2
    z12 = x12 y12 = (x2)6 (y3)4 = (u)6 (u)4 = u u = u


    z1 = xy
    z2 = y2
    z3 = x
    z4 = y
    z5 = xy2
    z6 = u
    z7 = xy
    z8 = y2
    z9 = x
    z10 = y
    z11 = xy2
    z12 = u

    This is what I have so far. Part 2 in my next reply.
  9. Nov 5, 2003 #8


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    Just a couple of observations:

    Once you've goten to zn=u, you're done, you'll only repeat.

    What you're describing is true for all x and y that commute (whenever xy=yx) there's no need to restrict yourself to abelian groups.
    Last edited: Nov 5, 2003
  10. Nov 5, 2003 #9
    Yes. I think I got this. Is this because the Group is finite?

    I am not sure what you are saying here:

    You are referring to my loose definition of the Commutative Property of the group correct?

    Could you explain the second point. If you feel you are simplifying it too much you probably are not.

    BTW NateTG, as far as I know we have yet to talk about isomorphism. If we had, I wasn't aware of it. But I don't recall a lecture on isomorphism as of yet.
  11. Nov 5, 2003 #10
    I'm not sure what I am supposed to do here:

    I am not sure of what to take permutations. Do I simply take any group, say S3, work out the permutations, then find a condition in which the permutations of S3 commute?

    BTW, after doing the first exercise, I noticed that the elements produced looked like the dihedral group D6. I'm not sure what that would mean in the context of the excerises you gave me Hurkyl.

    Another note: I am very appreciative of both the efforts of Hurkyl and NateTG. I know it can be frustrating trying to explain certain concepts to me, but if one has the patience, I will try my best to figure things out. Thankyou.
  12. Nov 6, 2003 #11


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    You might try a few different permutation groups, just in case S3 isn't representative of the general case. Varying the types of things over which you search is good practice in general.

    And, it turns out, that it is important in this case; S3 really isn't a very good representative of the general case. While you can find permutations that commute in S3, it probably won't be enough to suggest the general result. I would suggest considering permutations over 4 or more objects.

    Almost; it's because z has finite order.

    When studying complex numbers, did you ever have to do the fun problem of computing the value of i^2003? The trick is to decompose the exponent as:

    i^(2003) = i^(2000) * i^3 = (i^4)^500 * i^3 = 1^500 * i^3 = i^3 = -i

    The same principle is at work in groups. If x has finite order, that means that x^k = 1 for some nonzero value of k. Then, if we want to know what x^n is for n >= k (or n < 0), we can apply the above trick to reduce the problem to computing x^m for some m in the range [0, k).

    And, incidentally, you may or may not have seen the entire group. In particular, you have only touched the cyclic subgroup that is generated by z; in principle, there could be just about anything else in the rest of the group! For example, the group itself could be nonabelian, even though x and y commute. "Nonabelian group" just says that some elements don't commute, it doesn't say all elements don't commute.

    I did not intend to point you towards D6. I'm tired and my algebra book isn't handy so I may be wrong, but isn't D6 nonabelian? If so, then TMK the (sub)group I had you generate only bears a superficial resemblance to D6.

    But it's a good observation nonetheless. It is often useful to express a group in terms of its generators.
  13. Nov 6, 2003 #12
    I just finished reading your last post Hurkyl. It might take me a little time to absorb what you are saying. But I will address some points at this time.

    No I have not. Strangely enough, the last time I remember complex numbers was in high school which was many years ago. However I do see what you are suggesting.

    Right. I understand this.

    I believe it is. I am not sure but can the permutations of D6 be expressed as the symmetries of an equilateral triangle? If so, I don't think that D6 is abelian.

    I started out with S5 and picked two permutations of S5 of the order 3 and 2:

    (1 2 3)(4 5)

    I see that

    (1 2 3)(4 5) = (4 5)(1 2 3)

    Therefore these two cycles commute(?) Is this what you meant?

    It seems to me from the two permutations that I picked, the product of two disjoint cycles will always commute. Is this the condition that you were inferring in the second exercise?
  14. Nov 6, 2003 #13


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    Yep, that's the right condition! IIRC, that's also a necessary condition for two permutations to commute, but I don't remember for sure.

    So (1 2 3)(4 5) should have order 6. Using this fact, can you generate a subgroup of S5 of order 6?
  15. Nov 6, 2003 #14
    What is IIRC? I am not familiar with that abbreviation.

    I need to think on this one a bit more.
  16. Nov 6, 2003 #15


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    IIRC stands for something like "If I recall correctly"

    Hurkyl: Disjoint cycles is sufficient not necessary.
    For example:
    a=(1 2 3)(4 5)
    b=(1 2 3)

    commute (since b=a4), but are not disjoint cycles.

    But, thanks to you, I'm going to have to go home, and see if I can categorize all commutative subgroups of Si.


    D6 is not abelian (commutative).

    any finite group generated by a single element a is going to be of the form
    G = 1 a a2 a3 a4 a5...
    and since ax*ay=ax+y=ay*ax
    any group generated by a single element will be commutative.

    So, if you generated a subgroup using a single element and got D6 you made a mistake somewhere.
    Last edited: Nov 6, 2003
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