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For the group action of $G=S_3$ on $A=S_3$, written $g\cdot a=ga$, find the image of $(1\;2\;3)$ under the permutation representation $S_3\to S_A\cong S_6$.

Here's the given solution to the problem: Recall that the permutation representation of the group action is given by the map $$\Sigma: G\to S_A$$ $$g\mapsto \sigma_g, \sigma_g(a) = ga,$$ where $g\in G, a\in A$. Note that $A = \{(1), (1\;2),(1\;3), (2\;3), (1\;2\;3), (1\;3\;2)\}$. So, $$\begin{array}{|r|c|l|}\hline a & \sigma_{(1\;2\;3)} & \sigma_g(a) = ga\\ \hline(1) & \overset{\sigma_{(1\;2\;3)}}\mapsto & (1\;2\;3)(1) = (1\;2\;3) \\ \hline (1\;2) & \overset{\sigma_{(1\;2\;3)}}\mapsto & (1\;2\;3)(1\;2) = (1\;3) \\ \hline (1\;3) & \overset{\sigma_{(1\;2\;3)}}\mapsto & (1\;2\;3)(1\;3) = (2\;3) \\ \hline (2\;3) & \overset{\sigma_{(1\;2\;3)}}\mapsto & (1\;2\;3)(2\;3) = (1\;2) \\ \hline (1\;2\;3) & \overset{\sigma_{(1\;2\;3)}}\mapsto & (1\;2\;3)(1\;2\;3) = (1\;3\;2) \\ \hline (1\;3\;2) & \overset{\sigma_{(1\;2\;3)}}\mapsto & (1\;2\;3)(1\;3\;2) = (1). \\ \hline \end{array}$$

Observe that if we enumerate $A$ like so: $$A = \{\overset{\bf 1}{(1)}, \overset{\bf 2}{(1\;2)}, \overset{\bf 3}{(1\;3)}, \overset{\bf 4}{(2\;3)}, \overset{\bf 5}{(1\;2\;3)}, \overset{\bf 6}{(1\;3\;2)}\},$$ then the image of $(1\;2\;3)$ under the permutation representation $S_3\to S_A\cong S_6$ is determined by the following:

Starting at ${\bf 1}$ and consulting the mappings in the above table, we get $$({\bf1} \; {\bf5} \; {\bf6})({\bf2} \; {\bf3} \; {\bf4}).$$

Here's my question: Is this answer unique exactly? The problem statement says to find the image of $(1\;2\;3)$, but I'm slightly confused on the enumeration part. What stops anybody from just choosing the enumeration of $A$ to be $$A = \{\overset{\bf 1}{(1)}, \overset{\bf 2}{(1\;2)}, \overset{\bf 3}{(1\;3)}, \overset{\bf 5}{(2\;3)}, \overset{\bf 6}{(1\;2\;3)}, \overset{\bf 4}{(1\;3\;2)}\}?$$ In other words, perhaps I didn't think of the elements of $S_3$ in some kind of order. Because then we get the image to be $$({\bf 1\;6\;4})({\bf 2 \;5\;6}).$$

Am I missing something here? Apparently $S_A \cong S_6$, which means that the elements should map to "something" that have the same order (if I'm not mistaken), because then we wouldn't have an isomorphism. But how exactly do we know this ordering doesn't work beforehand? Can someone explain what my logical fallacy might be here?

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    $\begingroup$ If you didn't re-label $A$ then you'd get $$( (1),(123),(132))((12),(13),(23)).$$ For each way of labelling $A$ with the labels $1,2,3,4,5,6$ you get a different isomorphism between $S_A$ and $S_{\{1,2,3,4,5,6\}}=S_6$. $\endgroup$ – ancientmathematician Mar 8 '17 at 16:14
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Note that if you regard the Cayley map $S_3\to S(S_3)$, then $$(1\,2\,3\,)\mapsto((1)\,(1\,2\,3\,)\,(3\,2\,1))((1\,2)\,(1\,3)\,(2\,3))$$ as note in the comments. This would be the image of $(1\,2\,3)$ under the natural map.

As you observe, if you compose with an isomorphism $S(S_3)\to S_6$, uniqueness is only up to how you label elements of $S_3$ (and the only mistake you are making is thinking that there is some labelling that doesn't work). I personally like the labelling $$ \{\overset{\bf 1}{(1)}, \overset{\bf 2}{(1\;2\;3)}, \overset{\bf 3}{(3\;2\;1)}, \overset{\bf 4}{(1\;2)}, \overset{\bf 5}{(1\;3)}, \overset{\bf 6}{(2\;3)}\} $$ so that $(1\,2\,3\,)\mapsto(1\,2\,3)(4\,5\,6)$.

In general, this is the only freedom you have, but the case of $S_6$ is actually very special. For $S_6$, there is an exotic isomorphism $S_6\to S_6$ determined by the mapping \begin{align} (1\;2)&\mapsto(1\;2)(3\;4)(5\;6)\\ (2\;3)&\mapsto(1\;4)(2\;5)(3\;6)\\ (3\;4)&\mapsto(1\;3)(2\;4)(5\;6)\\ (4\;5)&\mapsto(1\;2)(3\;6)(4\;5)\\ (5\;6)&\mapsto(1\;4)(2\;3)(5\;6) \end{align} The composition $S_3\to S(S_3)\cong S_6\to S_6$ using my labelling above yields $$(1\;2\;3)\mapsto(2\;4\;6).$$

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