# Group Action of S3 on a set of ordered pairs

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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

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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 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).

Gold Member
Thanks Deveno

Really informative as usual!

Peter

Gold Member
Thanks

Peter