What are the conjugates to $(12)$ in the symmetric group $S_3$?
I know that in the symmetric group the elements are as follows: The identity, $(12)$, $(13)$, $(23)$, $(123)$, and $(132)$.
I want to find the conjugates of each of these with $(12)$. So I know you have to use the fact that if $h=(12)$ then $ghg^{-1}$ will get me my conjugate. However I'm confused as to how this happens.
How does $(132)(12)(132)^{-1}=(13)$? Because surely, $(132)^{-1}$ is $(231)=(123)$ Then $(132)(12)(123)$ is what I need to compute.
This is my mind goes as follows: Starting from RHS- 1 goes to 2 which goes to 1, so $(1)$ Then 2 goes to 3 which goes to 2 so $(132)(12)(123)=(1)(23)=(23)$?
I'm extremely confused and I just don't know how to work out these types of problems because I keep getting into trouble. Please someone help!
I can't also work out the conjugates of $(123)$.
$\endgroup$2 Answers
$\begingroup$Starting from the RHS, you have to go entirely to the left hand side. So for (132)(12)(123): 1 goes to 2, then 2 goes to 1, then 1 goes to 3, so $1\rightarrow 3$. Next 3 goes to 1, 1 goes to 2 and 2 goes to 1, so we have (13). You can now check that indeed: 2 goes to 3, 3 stays at 3, 3 goes back to 2. Conclusion: (132)(12)(123)=(13).
$\endgroup$ 3 $\begingroup$It is a fact that two permutations are conjugated if and only if they have the same cycle structure (found in many standard textbooks). See e.g.
In particular the conjugates of $(12)$ in $S_3$ are $(13)$ and $(23)$. Moreover $(123)$ is conjugated to $(132)$ in $S_3$ (note that they are not conjugated in $A_3$).
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