Find conjugacy classes of $S_3$

Question

I am trying to understand the best way to find the conjugacy classes of $S_3$. What I am doing to find these is first consider what elements would be in the center of the group $Z(S_3)$. Then, after that, I try finding the conjugacy classes for each element of $S_3$ that is not in the group. Namely, I try finding the conjugacy class of 2 cycles, so for $(12)$ and after that I try finding the conjugacy class for 3 cycles like $(123)$. Doing all the work of computing the conjugacies, I found taht the conjugacy class of $(12)$ is $\{(12),(13),(23)\}$ and the conjugacy class of $(123)$ is $\{(123),(132)\}$. Now my question is, is there a more efficient way to find these conjugacy classes, without computing all the conjugacies? For instance for the group $A4$, I think this approach is not efficient and definitely harder to apply. Thanks!

Answer 1

Yes there is a more efficient way, and your brute force calculation for $S_3$ hints at the answer, and how to find it.

Try conjugating a $2$ cycle in $S_4$ with various elements. You'll soon see the pattern and can guess the whole class. Then try finding a few conjugates and then the class of $(12)(345)$ in $S_7$.

Then you'll have a theorem to state and you'll know how to prove it.

(By the way, the center of the symmetric group is just the identity. Can you prove that?)

There's now a link to the duplicate of your question, with an answer. If you read that you'll miss the chance to discover it for yourself.