Finding conjugacy classes I've been having problems with finding conjugacy classes. I don't really understand how to do it properly. 
Say we look at a S3 group: $S_3=]e, (12), (13), (23), (123), (132)]$
If we look at just $(12)$ first:
$e(12)e^{-1} = e(12)e=(12)$ I can understand this.
$(12)(12)(12)^{-1} = (12)(12)(12) = (12)$ Now here I'm a little confused. Wouldn't $(12)^{-1} = (21)$?
Then $(13)(12)(13)^{-1} = (13)(12)(13) = (32)$ how did that become $(32)$? 
I thought you would start from the right so do $(12)(13)$ first then that would be $(123)$ then $(13)(123)$: 1 stays the same. 3 goes to 2? then 3 stays so (123)? I'm really confused
 A: (12) and (21) are the same permutation.
As for (13)(12)(13), well, simply compute what it does to 1,2,3:


*

*(13)(12)(13)1=(13)(12)3=(13)3=1

*(13)(12)(13)2=(13)(12)2=(13)1=3

*(13)(12)(13)3=(13)(12)1=(13)2=2


So (13)(12)(13) leaves 1 fixed, and swaps 2 and 3.
Also, (12)(13) is not (123). Check:


*

*(12)(13)1=(12)3=3

*(12)(13)2=(12)2=1

*(12)(13)3=(12)1=2


Thus, (12)(13) sends 1 to 3, and 3 to 2, and 2 to 1. So (12)(13)=(132), not (123).
A: $(12)$ and $(21)$ are the same permutation (which maps $1$ to $2$, $2$ to $1$, and leaves everything else unchanged).
Likewise, $(13)(12)(13) = (23)$ because it maps $2$ to $3$, $3$ to $2$ and leaves everything else unchanged (including $1$). Also, $(12)(13) = (132)$, not $(123)$.
A: Your confusion is coming from the composition of permutations. Here is an example
suppose we want to know $(1 \, 2 \, 3)( 1\, 3)$. Then we start from the rightmost cycle, i.e. $(1 \, 3)$. Here $1$ goes to $3$, then we consider the next cycle to the left, here $3$ goes to $1$, so we will conclude that the $1$ (from the rightmost cycle) goes to $1$ (because $1 \to 3 \to 1$).
Now we consider $3$ from the rightmost cycle, this $3$ goes to $1$, then in the left cycle $1$ goes to $2$, thus $3$ goes to $2$. So we have
$$(1 \, 2 \, 3)( 1\, 3)=(1)(3 \, 2)=(3 \, 2).$$
Once you get this then conjugacy will become easier to handle.
