1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You are in the right track, but you haven't justified the classes you list. First you claimed a class had 3 elements. Now you claim it has 2. You haven't actually shown this to be the case. Maybe instead of a class with two elements it is 2 with one each? $\endgroup$ – Andrés E. Caicedo Jan 5 at 15:26
  • $\begingroup$ In more detail, the issue is: you need to justify that things you list in the same class actually are conjugate. Maybe having the same cycle structure is not enough. In fact, you know it is not, since you separated one of the classes in your original list into two pieces. $\endgroup$ – Andrés E. Caicedo Jan 5 at 15:28
  • $\begingroup$ Yes that's true, at first I looked at sets of the form $\{xgx^{-1}~:~x\in S_4,~g\in G\}$ it should have been $x\in G$ $\endgroup$ – John Cataldo Jan 5 at 15:30
0
$\begingroup$

The list

  • $\{(1)\}$
  • $\{(1234),(1432)\}$
  • $\{(12)(34),(14)(32)\}$
  • $\{(13),(24)\}$
  • $\{(13)(24)\}$

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 $(14)(32)(1234)(14)(32)=(1432)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.