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According to the book Abstract Algebra, Theory and Applications by T. W. Judson (Example 8. Ch. 14) :

The nontrivial orbits of the conjugacy on $S_3$ are $${\{(1)}\},\ {\{(123), (132)}\},\ {\{(12), (13), (23)}\}. $$ So, the class equation is $6=1+2+3$.

The first wrong thing here is that ${\{(1)}\}$ is not nontrivial since it is singleton. The second wrong statement is that ${\{(123), (132)}\}$ and ${\{(12), (13), (23)}\}$ are not orbits at all. Consider $(12)$. Based on the definition, $O_{(12)}$ is the set of the elements such that there exist some $g\in G$ (and $g\ne (1)$) such that $g(12)=(12)g$. But there is no element like that and the book says $(12)$ and $(23)$ are; which means $(12)(13)=(13)(12)$ which is not correct. Same argument holds for ${\{(123), (132)}\}$.

Am I wrong or the book? and why?

Thanks a lot.

1 Answers1

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You are taking a wrong definition of the conjugacy. You say that in a group $G$ two elements $g_1,g_2$ are conjugated if there exists $x\in G$ such that $xg_1x^{-1} = g_2$. So in your case, $(12)$ and $(13)$ are conjugated because $(23)(12)(23) = (13)$, also $(12)$ and $(23)$ are conjugated because $(123)(12)(132) = (23)$. The same exercise can be done with the order 3 elements in $S_3$.

Those are the congujacy classes of $S_3$. You can prove in general the following statement:

Let $\sigma,\tau\in S_n$ and take the decomposition of disjoint cycles of them: $\sigma=c_1c_2\cdots c_r$ and $\tau=c'_1c'_2\cdots c'_s$. Then if $r=s$ and if you can reorder the cycles such that $length(c_i) = length(c'_i)$ for every $i$, then they are in the same conjugacy class.

Even more, in the case of $S_n$ there is a concrete algorithm to compute the $x$ of our definition.

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