Determination of character table of symmetric group:S3

This page describes various ways to determine the character table of symmetric group:S3. The idea is to determine as much as possible wit has little knowledge of the representation theory of these groups as we can manage, therefore making the discussion suitable for people who know only basic facts about the groups and basic facts of linear representation theory. For a detailed discussion of the character theory, see linear representation theory of symmetric group:S3.

We build on the basic information on conjugacy class structure available at element structure of symmetric group:S3.

Finding the degrees of irreducible representations using number-theoretic constraints
The following facts are known in general:


 * 1) Sum of squares of degrees of irreducible representations equals order of group: In this case, it tells us that the sum of the degrees of irreducible representations is 6, the order of the group.
 * 2) Number of irreducible representations equals number of conjugacy classes: In this case, it tells us that there are 3 irreducible representations, because there are 3 conjugacy classes.
 * 3) Degree of irreducible representation divides order of group: In this case, it tells us that the degrees of irreducible representations all divide 6, the order of the group.
 * 4) Number of one-dimensional representations equals order of abelianization: In this case, the derived subgroup is A3 in S3 (order three) and the abelianization (the quotient group) is isomorphic to cyclic group:Z2. So, the number 1 occurs two times as a degree of irreducible representation.
 * 5) Order of inner automorphism group bounds square of degree of irreducible representation

In fact, we do not need all these facts to determine the degrees of irreducible representations. Some combinations of subsets of these facts suffice. For instance:


 * (1) and (2) alone: By (2), there are three irreducible representations. Denote their (positive integer) degrees by $$d_1, d_2, d_3$$, with $$d_1 \le d_2 \le d_3$$. By (1), we have $$d_1^2 + d_2^2 + d_3^2 = 6$$, where $$d_1, d_2, d_3$$ are positive integers. The only solution to this system is $$d_1 = 1, d_2 = 1, d_3 = 2$$
 * (1) and (4) (even without knowledge of the precise form of the abelinization): The sum of squares of degrees of irreducible representations is 6. This opens up two possibilities: either there are six irreducible representations of degree one, or the degrees of irreducible representations are 1, 1, 2. Since the number of occurrences of 1 as a degree equals the order of the abelianization, and since the group is non-abelian, the first possibility is ruled out, so 1, 1, 2 is the only one.
 * (2), (4), and (5): (4) gives us that there are two irreducible representations of degree 1, and no more. (2) tells us that there is one more irreducible representation, so its degree must be more than 1. (5) tells us that the square of the degree is less than or equal to the order of the inner automorphism group, which is less than or equal to the order of the group, namely 6. In particular, 2 is the only natural number greater than 1 whose square is less than or equal to 6. Thus, 2 is the degree of the remaining irreducible representation.

Finding the character table
The focus here is on constructing the character table using minimal knowledge of the irreducible representations. This is to explain how to think about the problem even without having a clear idea of the specific story behind the group. We assume, however, that the conjugacy class structure is fully computed.

We have found that the degrees of irreducible representations are 1, 1, and 2. We know that the trivial representation (sending everything to the $$1 \times 1$$ matrix) is one of the irreducible one-dimensional representations. We do not know what the others are.

This is the current known character table:

Let's give names to the unknowns:

Let us first try to determine the second row. We know that both $$x$$ and $$y$$ are roots of unity (because the representation is one-dimensional). Since $$y$$ is the image of $$(1,2)$$, it must be either 1 or -1. Since $$x$$ is the image of an element of order three, it must be either 1 or a primitive cube root of unity.

Row orthogonality between the first two rows gives us:

$$ 1+ 2x + 3y = 0$$

Since $$y$$ is rational, this forces $$x$$ to be rational, so $$x = 1$$. Thus, $$y = -1$$, and we get:

We can now use column orthogonality to find $$z$$ and $$w$$. Column orthogonality between the first and second column gives us:

$$1 + 1 + 2z = 0$$

This solves to $$z = -1$$.

Column orthogonality between the first and third column gives us:

$$1 + (-1) + 2w = 0$$

This solves to $$w = 0$$.

Plug these values in and obtain the full character table: