Find all conjugacy classes in \(S_3\), listing the elements of each class.
Solution
1. Conjugation in \(S_3\) preserves cycle type.
For \(\sigma\in S_3\) and a cycle \[ \tau=(i_1\,i_2\,\ldots\,i_r), \] one has \[ \sigma\tau\sigma^{-1} = (\sigma(i_1)\,\sigma(i_2)\,\ldots\,\sigma(i_r)). \] Thus conjugating merely relabels the entries of a cycle and hence preserves its cycle type.
2. The identity forms the conjugacy class\[ \{1\}. \]
For every \(\sigma\in S_3\), \[ \sigma1\sigma^{-1}=1. \]
3. The three transpositions form one conjugacy class:\[ \{(12),(13),(23)\}. \]
By step 1, every conjugate of a transposition is again a transposition. Conversely, for any two transpositions \[ (ij) \qquad\text{and}\qquad (rs), \] choose \(\sigma\in S_3\) with \[ \sigma(i)=r, \qquad \sigma(j)=s. \] Then \[ \sigma(ij)\sigma^{-1}=(rs). \] Thus all three transpositions are conjugate.
4. The two \(3\)-cycles form one conjugacy class:\[ \{(123),(132)\}. \]
By step 1, conjugates of a \(3\)-cycle are \(3\)-cycles. Moreover, \[ (23)(123)(23)^{-1} = (132), \] so the two \(3\)-cycles are conjugate.
5. These are all conjugacy classes of \(S_3\).
The elements of \(S_3\) are exactly \[ 1,\ (12),\ (13),\ (23),\ (123),\ (132). \] Steps <1>2–<1>4 partition these six elements into three conjugacy classes. Hence no further classes exist.
6. Therefore the conjugacy classes are\[ \boxed{ \{1\}, \qquad \{(12),(13),(23)\}, \qquad \{(123),(132)\}. } \]
This is the class decomposition established in steps <1>2–<1>5.
7. Q.E.D.
Step 6 is the requested list.