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I am trying to find the centraliser of $(12)(34)$ in $S_4$.

I have that

$$\begin{align}C_{S_4}((12)(34)) &=\{g \in G : g(12)(34)g^{-1}=(12)(34) \} \\ &= \{g \in G : (g(1) g(2))(g(3)g(4))=(12)(34) \} \end{align}$$

Apparently the answer is $\{e, (12),(34), (12)(34),(1324),(1423),(13)(24),(14)(23) \}$ See http://www.maths.manchester.ac.uk/~rs/AlgStSol7A.pdf

I just cannot see how $(1324)$ and $(13)(24)$ can be part of the centraliser as they seem to violate $g(1) \in \{1,2\}, g(2) \in \{1,2\}$ and $g(3) \in \{3,4 \}, g(4) \in \{3,4 \}$

I was given the hint: Use the orbit-stabiliser. But I cannot see how this is any practical use

Trajan
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    Hint: You could have $(g(1)g(2)) = (34)$ and $(g(3)g(4)) = (12),$ for example. – Geoff Robinson Feb 02 '15 at 18:36
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    That condition is not what you want. Note that $(1324)$ sends $1$ to $3$ and $2$ to $4$, while it sends $3$ to $2$ and $4$ to $1$; so indeed conjugating $(13)(24)$ by this leaves $(13)(24)$. – Pedro Feb 02 '15 at 18:36

2 Answers2

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I know it has been answered, but i will give an algorithm to find explicitly those permutations.

Observe that the result of the conjugation by $\sigma$ in the centralizer may give $(12)(34)$ written in a different but equivalent way, with its integers and cycle order interchanged (in fact, the only permutation in the centralizer which does not change the way of expression is $(1)$ ).

For example, $\sigma = (13)(24)$ in the centralizer gives us $\sigma(12)(34) \sigma^{-1} = (34)(12)$ (which is indeed equal to $(12)(34)$ ) when calculating via the application of $\sigma$ to the integers of $(12)(34)$.

Using this idea, we just have to find then all the equivalent ways to express $(12)(34)$. Those ways are exactly:

$$(12)(34),(12)(43),(21)(34),(21)(43),(34)(12),(43)(12),(43)(21),(34)(21)$$

which are $8$ in total. Now to find those permutations you can simply use the fact that the result of the conjugation is just the application of $\sigma$ to the integers of $(12)(34)$.

For example, to find for $(43)(12)$, $\sigma$ must satisfy:

$\sigma(1)=4,\sigma(2)=3,\sigma(3)=1,\sigma(4)=2;$ which is $\sigma =(1423). $

user2345678
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Hint: The centraliser of $(12)(34)$ in $S_4$ is given by what you have shown:

$$C_{S_4}((12)(34))=\{g\in G| g(12)(34)g^{-1}=(12)(34)\}$$

Firstly we note that by cycle type, conjugating it with anything will give you back a $2-2$ cycle type(since these are all of the things in the conjugacy class), so now we just need to find what doesn't shift it to one of the other elements of the conjugacy class, i.e. $(13)(24)$ and $(14)(23)$.

Permute
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