Dummit and Foote Section 4.1 Group Actions and Permutation Representations, Exercise 4 (first part of exercise) reads:(adsbygoogle = window.adsbygoogle || []).push({});

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]

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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)}

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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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# Group Action of S3 on a set of ordered pairs

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