Let $G=\{(1),(1234),(13)(24),(12)(34),(14)(32),(1432),(13),(24)\}\le S_4$ It is a subgroup and $Z(G)=\{(1),(13)(24)\}$
Find all the conjugacy classes of $G$.
I was surprised because I didn't use the fact that they give us $Z(G)$.
But here is how I think I did it:
We know that for a transposition $\tau$ and a cycle $\sigma=(a_1...a_n)$ $$\tau\sigma\tau^{-1}=(\tau(a_1)...\tau(a_n))$$ ands since any permutation can be written as a product of cycles or a product of transpositions and using the fact that if $c_i,~i\in\{1,...,n\}$ are cycles we can rewrite $$\tau (c_1...c_n)\tau^{-1}$$ as $$\tau c_1\tau^{-1}\tau c_2\tau^{-1}...\tau c_n\tau^{-1}$$ we can deduce that the action of conjugation preserves the structure of a permutation.
So the classes of permutations are made of elements that have the same structure (same kind of decomposition into cycles, like $(ab)(cd)$, $(ac)(bd)$ and $(ad)(bc)$ are all in the same class for example).
So we have the following classes:
- $\{(1)\}$
- $\{(1234),(1432)\}$
- $\{(12)(34),(13)(24),(14)(32)\}$
- $\{(13),(24)\}$
So they probably gave us $Z(G)$ for a reason, but I don 't know where I should have used it...
Edit: Actually since $(13)(24)\in Z$ it should be a separate class. So instead of having the class $\{(12)(34),(13)(24),(14)(32)\}$ it should be broken into two classes $\{(12)(34),(14)(32)\}$ , $~\{(13)(24)\}$. So I guess this is where we have to use that information...
So now would you agree with those classes and the way I justified them?
The list
Is a good candidate but it could be that some permutations are conjugate to each other by elements not in $G$. And that would prevent them from sharing a class.
So verifying if there are elements $\sigma$ in $G$ that send an element of a class to another of the same class doesn't take that long. For example $(1432)(13)(1234)=(24)$ so $\{13),(24)\}$ is a class of $G$ because $(1432)\in G$
We also get $(13)(12)(34)(13)=(14)(32)$ and $(14)(32)(1234)(14)(32)=(1432)$