Group Action of S3 on a set of ordered pairs

[Math Amateur]

Dummit and Foote Section 4.1 Group Actions and Permutation Representations, Exercise 4 (first part of exercise) reads:

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

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So, S_3 = { 1, (2 3), (1 3), (1 2), (1 2 3), (1 3 2) }

Now my first calculations follow:

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

Thus calculating elements of this orbit:

(1)\star(1,1) = {\sigma}_1(1,1) = ({\sigma}_1(1), {\sigma}_1(1)) = (1,1)

(2 3)\star(1,1) = {\sigma}_{23}(1,1) = ({\sigma}_{23}(1), {\sigma}_{23}(1)) = (1,1)

(1 3)\star(1,1) = {\sigma}_{13}(1,1) = ({\sigma}_{13}(1), {\sigma}_{13}(1)) = (3,3)

(1 2)\star(1,1) = {\sigma}_{12}(1,1) = ({\sigma}_{12}(1), {\sigma}_{12}(1)) = (2,2)

(1 2 3)\star(1,1) = {\sigma}_{123}(1,1) = ({\sigma}_{123}(1), {\sigma}_{123}(1)) = (2,2)

(1 3 2)\star(1,1) = {\sigma}_{132}(1,1) = ({\sigma}_{132}(1), {\sigma}_{132}(1)) = (3,3)

Thus the orbit of S_3 = {(1.1), (2,2), (3,3)}

Next orbit:

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

(1)\star(1,2) = {\sigma}_1(1,2) = ({\sigma}_1(1), {\sigma}_1(2)) = (1,2)

(2 3)\star(1,1) = {\sigma}_{23}(1,2) = ({\sigma}_{23}(1), {\sigma}_{23}(2)) = (1,3)

(1 3)\star(1,2) = {\sigma}_{13}(1,2) = ({\sigma}_{13}(1), {\sigma}_{13}(2)) = (3,2)

(1 2)\star(1,2) = {\sigma}_{12}(1,2) = ({\sigma}_{12}(1), {\sigma}_{12}(2)) = (2,1)

(1 2 3)\star(1,2) = {\sigma}_{123}(1,2) = ({\sigma}_{123}(1), {\sigma}_{123}(2)) = (2,3)

(1 3 2)\star(1,2) = {\sigma}_{132}(1,2) = ({\sigma}_{132}(1), {\sigma}_{132}(2)) = (3,1)

Thus the orbit of S_3 containing (1,2) = {(1,2), (1,3), (2,1), (2,3), 3,1), 3,2)}

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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

 

[micromass]

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

 

[Deveno]

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).

 

[Math Amateur]

Thanks Deveno

Really informative as usual!

Peter

 

[Math Amateur]

Thanks

Peter