I am considering the permutation representation of $S_4$ acting on $\{1,2,3,4\}$ with character $\chi_{\pi}$. The answer is $\chi_{\pi}=(4,2,1,0,0)$

Now I know that for the permutation representation $$\begin{align}\chi_{\pi} &= \text{ the number of fixed points of g on X} \\ &= \{x \in G \mid g \cdot x= x \} \end{align}$$

The conjugacy classes of $S_4$ are, just denoted by a representative, $e,(12),(123),(1234),(12)(34)$.

For the case of e it is easy and the answer will be $4$. I am struggling to get a good method (ie better than trial and error) for the other cases. For instance for the case $(12)$ we have $$ \{x \in G \mid g \cdot x= x \}= \{x \in G \mid g (12)= (12) \}$$ but I dont see how the g(12) can be worked with in an efficient way. Maybe it would help if there was some relation (for the symmetric group such that $g(12)=(f_1(1),f_2(2))$

  • $\begingroup$ $(1,2)$ replace $1$ and $2$ and therefore $3,4$ are fixed. $\endgroup$ – Ofir Schnabel Apr 29 '15 at 9:55

Hint : take a basis $(e_1,e_2,e_3,e_4)$. Then, for example, $\sigma$ sends $e_1$ to $e_{\sigma(1)}$. All you have to do to find $\chi(\sigma)$ is to compute the trace of the matric of $\sigma$ in this basis.

Let me take the example of $\sigma = (12)(34)$ and let $g$ be the morphism associated with $\sigma$. Then, $$g(e_1) = e_2 $$ $$g_(e_2) = e_1 $$ $$g(e_3) = e_4 $$ $$g(e_4) = e_3 $$ Therefore, the matrix of $g$ will be

$$\begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\0&0&0&1 \\0&0&1&0 \end{pmatrix}$$ It is obvious that $\mathrm{tr}(g) = 0$.

Note that computing the trace of a morphism is generally very simple and is an efficient way of computing characters of a representation : all you have to do is to choose a suitable representant of the conjugacy class and a suitable basis to compute the trace.

  • $\begingroup$ Thanks. Really good method! $\endgroup$ – Permian Apr 29 '15 at 12:10

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