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My book has an example that a permutation group $S_n$ acts on itself through conjugation,i.e. the group action is defined by $\tau x\tau^{-1}$ that $x$ and $\tau$ are both elements of $S_n$.

Why the order of multiplying $\tau$ and $\tau^{-1}$on $x$ matters? For $\tau x\tau^{-1}$, I can prove it is a group action:
1.$exe^{-1}=x$
2.$(g_1 g_2)x = g_1 g_2 x g_2^{-1} g_1^{-1} = g_1(g_2 x g_2^{-1})g_1^{-1} = g_1(g_2(x)) $

But for $\tau^{-1} x \tau$,
$(g_1 g_2)x = g_2^{-1} g_1^{-1} x g_1 g_2 $ while $(g_1(g_2(x)) = g_1(g_2^{-1}x g_2) = g_1^{-1}g_2^{-1}xg_2g_1$

However, if $\tau y\tau^{-1}=x$ , then $\tau^{-1} x \tau = y$, so isn't it just an action in the opposite direction? Should $\tau^{-1} x \tau$ be considered a group action?

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  • $\begingroup$ Where did you read that “group action of a symmetric group is the composition of function”. Note that I don't disagree. I just want to be aware of the context. $\endgroup$ – José Carlos Santos May 27 '18 at 8:41
  • $\begingroup$ en.wikipedia.org/wiki/Symmetric_group here "the symmetric group defined over any set is the group whose elements are all the bijections from the set to itself, and whose group operation is the composition of functions." $\endgroup$ – loct May 27 '18 at 8:51
  • $\begingroup$ Oh no I mixed it up...feel sorry about this.. $\endgroup$ – loct May 27 '18 at 8:54
  • $\begingroup$ Note that the passage you quote is about “group operation”, not about “group action”. $\endgroup$ – José Carlos Santos May 27 '18 at 8:54
  • $\begingroup$ The part about what I read from wiki has been deleted, in order not to confuse readers... $\endgroup$ – loct May 27 '18 at 9:02
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It turns out that the second action is not an action of the symmetric group $S_n$, but an action of its opposite group: $(S_n,\bullet)$, with $g\bullet h=h.g$. Since the symmetric group is not commutative, these actions are not the same.

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