I’m trying to learn a little bit about representation theory because it sometimes comes up in discussions in some math hobby groups that I’m in. I have no background in representation theory and my question is really, really naive.

I don’t know how to compute a character table for a small, easy-to-work-with group by hand. Is there an analog to a character table for permutation representations and is it simpler to compute by hand?

Richard Borcherds’ first lecture on representation theory introduces some of the basic concepts and constructs a character table for $S_3$ at 20m25s.

$$ \begin{array}{lccc} & \begin{matrix} \\ \\ \mathrm{id} \end{matrix} & \begin{matrix} (12) \\[-0.2em] (23) \\[-0.2em] (31) \end{matrix} & \begin{matrix} \\ (123) \\[-0.2em] (132) \end{matrix} \\[0.5em] \hline \chi_1 & 1 & \phantom{-}1 & \phantom{-}1 \\ \chi_2 & 1 & -1 & \phantom{-}1 \\ \chi_3 & 2 & \phantom{-}0 & -1 \end{array} $$

And then he says some facts about this table that I find absolutely amazing:

  1. It’s square.
  2. The columns are orthogonal.
  3. The rows are orthogonal (when weighted by conjugacy class size).

I didn’t get this at first, but $\chi_3$ corresponds to the standard representation, which only costs you two dimensions in $\mathbb{C}$ to represent (as opposed to 3 if you do it naively by translating permutations into matrices).

What follows is my attempt to build a permutation character table for $S_3$.

Wikipedia’s entry on character tables mentions that the table itself encodes information about irreducible representations over $\mathbb{C}$. However, you can take traces of matrices in any vector space, this makes me wonder whether it is possible to construct something analogous to this table for permutation representations of $S_3$.

This would be a binary matrix, since traces would give us $1$ or $0$ and would count the parity of the number of fixed points.

I’ll call my new pseudo-characters $\psi_1, \psi_2, \dotsc$

We have the identity pseudocharacter $\psi_1$, which sends everything to $1$, associated with the trivial permutation representation.

We have the representation that sends elements of the transposition conjugacy class to $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ and everything else to the identity matrix. This representation is associated with a pseudocharacter $\psi_2$ which sends everything to $0$ because it’s two-dimensional and two is even. I guess this thing is irreducible as a permutation representation, but I don’t know how to prove it one way or the other. It certainly is not irreducible as a complex representation, though, since it isn’t listed in the above character table.

And finally we have the faithful permutation representation, which we use in the above table to name the elements of the conjugacy classes of $S_3$ above. The identity permutation has three fixed points, a transposition has one, and derangements have no fixed points, so $\psi_3$ sends the identity and the transpositions to $1$ and the others to $2$.

We end up with the following table, which is much less nice.

$$ \begin{array}{lccc} & \begin{matrix} \\ \\ \mathrm{id} \end{matrix} & \begin{matrix} (12) \\[-0.2em] (23) \\[-0.2em] (31) \end{matrix} & \begin{matrix} \\ (123) \\[-0.2em] (132) \end{matrix} \\[0.5em] \hline \psi_1 & 1 & 1 & 1 \\ \psi_2 & 0 & 0 & 0 \\ \psi_3 & 1 & 1 & 0 \end{array} $$

We can also count the number of fixed points instead, which is also invariant under conjugation.

$$ \begin{array}{lccc} & \begin{matrix} \\ \\ \mathrm{id} \end{matrix} & \begin{matrix} (12) \\[-0.2em] (23) \\[-0.2em] (31) \end{matrix} & \begin{matrix} \\ (123) \\[-0.2em] (132) \end{matrix} \\[0.5em] \hline \psi_1 & 1 & 1 & 1 \\ \psi_2 & 2 & 0 & 2 \\ \psi_3 & 3 & 1 & 0 \end{array} $$

Anyway, since our group $S_3$ has three normal subgroups and we are effectively counting the number of fixed points in permutation representations of quotient groups of $S_3$, I think we can use the above table to tell us what kinds of numbers of fixed points for each conjugacy class are possible in any permutation representation of $S_3$, for example $\begin{bmatrix} 4 & 2 & 1 \end{bmatrix}$ is possible if we act on three elements standardly and leave a fourth untouched.

  • 4
    $\begingroup$ I don't really understand what you're trying to do. Why are you only taking the trace $\bmod 2$? You can just take the trace over $\mathbb{Z}$ which records the number of fixed points. This tells you the character of the permutation representation but you don't get a precise analog of the character table: 1) it's not square (typically there are many more permutation representations than conjugacy classes), and 2) the character is not a complete invariant; it's possible for two permutation representations to be non-isomorphic (as permutation representations) but have the same character. $\endgroup$ Jan 15 at 5:31
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    $\begingroup$ Maybe you would be interested in the table of marks? $\endgroup$
    – anon
    Jan 15 at 5:32
  • $\begingroup$ @QiaochuYuan I don't know what I'm doing. I was trying to pick a field besides $\mathbb{C}$ to build the analog of the character table in, and I went with the field with two elements, that's where the trace mod two comes from (since that's the trace in $\mathbb{F}_2^n$). It sounds like $\mathbb{C}$ is so special (for reasons I don't understand yet) that a table that you build based on a different field will simply lack the properties that make the character table interesting / a good description of the behavior of the group. $\endgroup$ Jan 15 at 5:48
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    $\begingroup$ Well, it's tricky. There is a character table of irreducible representations over a field $k$ but the basic theorems over $\mathbb{C}$ will only generalize if 1) the characteristic of $k$ does not divide $|G|$ (so working over $\mathbb{F}_2$ here is tricky since $2 \mid 6$) and 2) $k$ is what is called a "splitting field" of $G$; a sufficient condition here is that $k$ is algebraically closed. But again it just depends on what you're trying to do. $\endgroup$ Jan 15 at 5:54
  • $\begingroup$ @runway44, thanks for the reference. $\endgroup$ Jan 15 at 6:06

1 Answer 1


Every group action is the “union” of transitive actions. Every transitive action of a group $G$ is equivalent to the action on the set of (left) cosets $G/H$ for some $H\le G$ by (left) multiplication. The transitive permutation characters have the form $(1_H)^G$ where $1_H$ is the trivial character of $H$ and $(1_H)^G$ is the induced character of degree $|G:H|$. For $G=S_3$ we get the following table: $$\begin{pmatrix} 1&1&1\\ 2&0&2\\ 3&1&0\\ 6&0&0 \end{pmatrix}$$


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