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Group Action of S3 on a set of ordered pairs

  1. Oct 28, 2011 #1
    Dummit and Foote Section 4.1 Group Actions and Permutation Representations, Exercise 4 (first part of exercise) reads:

    Let [itex] S_3 [/itex] act on the set [itex] \Omega [/itex] of ordered pairs: {(i,j) | 1≤ i,j ≤ 3} by σ((i,j)) = (σ(i), σ(j)). Find the orbits of [itex] S_3 [/itex] on [itex] \Omega [/itex]

    ================================================================

    So, [itex] S_3 [/itex] = { 1, (2 3), (1 3), (1 2), (1 2 3), (1 3 2) }

    Now my first calculations follow:

    The orbit of [itex] S_3 [/itex] containing (1,1) = {g[itex] \star [/itex](1,1) | g [itex] \in [/itex] [itex] S_3 [/itex]}

    Thus calculating elements of this orbit:

    (1)[itex] \star [/itex](1,1) = [itex] {\sigma}_1[/itex](1,1) = ([itex] {\sigma}_1[/itex](1), [itex] {\sigma}_1[/itex](1)) = (1,1)

    (2 3)[itex] \star [/itex](1,1) = [itex] {\sigma}_{23}[/itex](1,1) = ([itex] {\sigma}_{23}[/itex](1), [itex] {\sigma}_{23}[/itex](1)) = (1,1)

    (1 3)[itex] \star [/itex](1,1) = [itex] {\sigma}_{13}[/itex](1,1) = ([itex] {\sigma}_{13}[/itex](1), [itex] {\sigma}_{13}[/itex](1)) = (3,3)

    (1 2)[itex] \star [/itex](1,1) = [itex] {\sigma}_{12}[/itex](1,1) = ([itex] {\sigma}_{12}[/itex](1), [itex] {\sigma}_{12}[/itex](1)) = (2,2)

    (1 2 3)[itex] \star [/itex](1,1) = [itex] {\sigma}_{123}[/itex](1,1) = ([itex] {\sigma}_{123}[/itex](1), [itex] {\sigma}_{123}[/itex](1)) = (2,2)

    (1 3 2)[itex] \star [/itex](1,1) = [itex] {\sigma}_{132}[/itex](1,1) = ([itex] {\sigma}_{132}[/itex](1), [itex] {\sigma}_{132}[/itex](1)) = (3,3)

    Thus the orbit of [itex] S_3 [/itex] = {(1.1), (2,2), (3,3)}

    Next orbit:

    The orbit of [itex] S_3 [/itex] containing (1,2) = {g[itex] \star [/itex](1,2) | g [itex] \in [/itex] [itex] S_3 [/itex]}

    (1)[itex] \star [/itex](1,2) = [itex] {\sigma}_1[/itex](1,2) = ([itex] {\sigma}_1[/itex](1), [itex] {\sigma}_1[/itex](2)) = (1,2)

    (2 3)[itex] \star [/itex](1,1) = [itex] {\sigma}_{23}[/itex](1,2) = ([itex] {\sigma}_{23}[/itex](1), [itex] {\sigma}_{23}[/itex](2)) = (1,3)

    (1 3)[itex] \star [/itex](1,2) = [itex] {\sigma}_{13}[/itex](1,2) = ([itex] {\sigma}_{13}[/itex](1), [itex] {\sigma}_{13}[/itex](2)) = (3,2)

    (1 2)[itex] \star [/itex](1,2) = [itex] {\sigma}_{12}[/itex](1,2) = ([itex] {\sigma}_{12}[/itex](1), [itex] {\sigma}_{12}[/itex](2)) = (2,1)

    (1 2 3)[itex] \star [/itex](1,2) = [itex] {\sigma}_{123}[/itex](1,2) = ([itex] {\sigma}_{123}[/itex](1), [itex] {\sigma}_{123}[/itex](2)) = (2,3)

    (1 3 2)[itex] \star [/itex](1,2) = [itex] {\sigma}_{132}[/itex](1,2) = ([itex] {\sigma}_{132}[/itex](1), [itex] {\sigma}_{132}[/itex](2)) = (3,1)

    Thus the orbit of [itex] S_3 [/itex] containing (1,2) = {(1,2), (1,3), (2,1), (2,3), 3,1), 3,2)}

    ===================================================================

    Could someone please indicate to me that the above calculations are proceeding correctly ( I am a math hobbyist working alone so I would appreciate someone indicating that my approach is correct)

    Dummit and Foote mention that a group acting on a set A partitions that set into disjoint equivalence classes under the action of G

    I was somewhat alarmed that I have, for the orbits, one set of 3 elements and another set of 6 elements.Previously I was under the impression (delusion??) that an equivalence relation partitioned a set into equal equivalence classes. The above tells me that the equivalence classes do not have to have the same number of elements - is that correct - or are my calculations of the orbits wrong?

    Peter
     
  2. jcsd
  3. Oct 28, 2011 #2

    micromass

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    Everyhting here is correct!!

    And indeed, the equivalence classes of a relation do not need to partition a set in sets of equal size. I'm glad to free you from this delusion :smile:
     
  4. Oct 28, 2011 #3

    Deveno

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    the idea you had (of the equivalence classes being the same size) is only true for a group congruence (equivalence in G "mod H").

    what happens is that the presence of a group multiplication on a set, introduces a certain kind of regularity on that set. for an ordinary set (with no group product), equivalence classes can be any size. imagine we have the following function:

    f: A --->A, where A = {a,b,c}, given by:

    f(a) = a
    f(b) = a
    f(c) = b

    we can define a relation on A by:

    x~y if f(x) = f(y).

    note that x~x since f(x) = f(x), and if x~y, then f(x) = f(y), so f(y) = f(x), so y~x, and finally, if x~y, and y~z, then f(x) = f(y), and f(y) = f(z), so f(x) = f(z), and thus x~z.

    so ~ is indeed an equivalence relation, but:

    [a] = {a,b} [c] = {c} (a and b share the same image under f, namely a, but c is the only element in A with f(x) = b), and the equivalence classes are not the same size.

    the above example is not a far-fetched one, many equivalence relations on a set arise in just such a way (and in fact, given an equivalence relation on a set, one can actually construct a function which produces that same partition). for your set Ω, we can do this like so:

    f(1,1) = (1,1)
    f(2,2) = (1,1)
    f(3,3) = (1,1)
    f(1,2) = (1,2)
    f(1,3) = (1,2)
    f(2,3) = (1,2)
    f(2,1) = (1,2)
    f(3,1) = (1,2)
    f(3,2) = (1,2)

    f is clearly a function Ω→Ω, and the relation (a,b)~(c,d) if f(a,b) = f(c,d) gives the same equivalence classes as your orbits.

    (very nice post, by the way, so clear).
     
  5. Oct 29, 2011 #4
    Thanks Deveno

    Really informative as usual!

    Peter
     
  6. Oct 29, 2011 #5
    Thanks

    Peter
     
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