Determining the character table of a finite group

This survey article discusses how to construct the survey article about::character table of a finite group. Some of the features of the article are:


 * The article focuses on constructing the character table even in the absence of complete information about the group -- for instance, if only some basic facts are known about the group, but its entire multiplication table and subgroup structure are unclear.
 * The article includes methods to construct characters without knowing the linear representations they come from.
 * Many of these methods work over fields of prime characteristic, where the prime does not divide the order of the group, as long as the field is a splitting field for the group.

Related survey articles

 * Finding linear representations of a group
 * Using the character table to find group-theoretic invariants

Writing the character table
The character table is a square matrix where the columns are indexed by conjugacy classes and the rows are indexed by representations. If the structure of the group is clearly understood, then the columns can easily be specified, which means that we know the number of rows as well as the number of columns. For later computations, it is also important to know the size of each conjugacy class.

By default, the left-most column is chosen for the conjugacy class of the identity element.

Linear characters
Linear characters are one-dimensional representations, or equivalently, characters of one-dimensional representations. These are homomorphisms from the group to the multiplicative group of the complex numbers. By obvious considerations, the image of a linear character is a cyclic group.

The principal character
The principal character, or trivial character, sends every element to the identity matrix of degree one. This is typically written in the first row, so the first row is all $$1$$s.

Other linear characters
To determine other linear characters, we need to know what the abelianization of the group $$G$$ is. Let's first discuss the special case where $$G$$ is a cyclic group.

If $$G$$ is cyclic of order $$d$$, then the linear characters of $$G$$ are given by maps sending the generator of $$G$$ to various $$d^{th}$$ roots of unity. There are $$d$$ such maps (one of which corresponds to the principal character). Suppose we write $$G$$ as $$\mathbb{Z}/d\mathbb{Z}$$. Then, the $$i^{th}$$ map sends $$j \in G$$ to the complex number $$e^{2\pi ij/d}$$.

If $$G$$ is a finite abelian group, then by the classification of finite abelian groups, $$G$$ is a direct product of cyclic groups. The linear characters of $$G$$ are all possible products of linear characters of each of the direct factors for any chosen direct decomposition of $$G$$.

If $$G$$ is not abelian, the linear characters of $$G$$ are precisely the linear characters of the abelianization of $$G$$, i.e., the quotient of $$G$$ by its commutator subgroup $$G'$$. For every linear character of $$G/G'$$, the corresponding linear character of $$G$$ sends every element of $$G$$ to the value of the linear character on its coset for the subgroup $$G'$$.

Using existing characters to find more characters
The number of linear characters equals the index of the commutator subgroup, or the order of the abelianization. This is less than the total number of conjugacy classes if the group is non-abelian, so there exist more characters. Here ,we discuss a few methods to find more characters using existing ones.

Getting off the ground: finding a character of degree two or higher
Determining whether a group has a character of degree two is equivalent to determining whether it has a quotient group that is isomorphic to a subgroup of $$U(2, \mathbb{C})$$. For this, it is useful to consider the classification of finite subgroups of U(2,C), and then check whether any of the quotients of the group is one of these.

Other methods include using some natural action that we know the group comes equipped with to obtain an irreducible representation.

Multiplying characters with linear characters
The product of an irreducible character with a linear character again yields an irreducible character of the same degree. The new character need not be distinct from the old character, but it often is.

Using outer automorphisms
Another way of getting new characters from existing ones is to use the outer automorphisms of a group that are not class-preserving. If $$\sigma$$ is an automorphism of a group $$G$$, and $$\chi$$ is an irreducible character, $$\sigma \circ \chi$$ is also irreducible. Further, if $$\sigma$$ is not class-preserving, it is possible that $$\sigma \circ \chi$$ is distinct from $$\chi$$.

More generally, if a character has kernel $$N$$ (the kernel can be determined as the union of all conjugacy classes where the character value equals that at the identity matrix) any automorphism of the quotient group $$G/N$$ can be used.

Multiplying two characters
The product of the characters of two (not necessarily distinct) linear representations is the character of a linear representation: the tensor product of the original representations. However, the tensor product of irreducible linear representations is not in general irreducible.

Nonetheless, since taking the product of two characters gives us a new (not necessarily irreducible) character, we can use techniques discussed later for decomposing characters to find irreducible constituents of this character that are not already in the table.

Basic methods
The degree of an irreducible representation is its dimension, which is also the character value at the identity. In particular, the first column of the character table lists the degrees of irreducible representations. There are very strong constraints on these degrees, such as:


 * Sum of squares of degree of irreducible representations equals order of group
 * Degree of irreducible representation divides index of abelian normal subgroup (in particular, degree of irreducible representation divides order of group
 * Order of inner automorphism group bounds square of degree of irreducible representation

In particular, if we know that there exists an abelian normal subgroup of large order, we get strong constraints on the degree. Even in the absence of this, the sum of squares condition is fairly strong, combined with the fact that we already know that the number of times $$1$$ occurs is the order of the abelianization.

Some examples
Suppose $$G$$ is a non-abelian group of order $$p^3$$. From some elementary group theory, we can deduce that:


 * The center of $$G$$ has order $$p$$: It is nontrivial because prime power order implies not centerless, and it cannot have order $$p^2$$ because then the quotient would be cyclic, and cyclic over central implies abelian.
 * The quotient of $$G$$ by its center is abelian, so the commutator subgroup is contained in the center. Hence, the commutator subgroup also has order $$p$$ and the abelianization has order $$p^2$$.
 * $$G$$ has an abelian normal subgroup of order $$p^2$$: Any subgroup of order $$p^2$$ is normal (nilpotent implies every maximal subgroup is normal) and any group of order $$p^2$$ is abelian. Finally, subgroups of order $$p^2$$ do exist.

Thus, from the general facts, $$G$$ has $$p^2$$ irreducible representations of degree one, and all the remaining irreducible representations must have degree $$p$$. Suppose there are $$k$$ representations of degree $$p$$. Then:

$$p^2 + kp^2 = p^3$$.

This solves to $$k = p - 1$$. Thus, there are $$p^2$$ irreducible representations of degree one and $$p-1$$ irreducible representations of degree $$p$$.

We were able to deduce all this without even knowing $$G$$ uniquely up to isomorphism. In fact, for all primes $$p$$, there are two isomorphism classes of such groups. For $$p = 2$$, there are dihedral group:D8 and quaternion group, and for odd $$p$$, there is prime-cube order group:U3p and prime-cube order group:p2byp.

As another example, suppose $$G$$ is a group of order $$48$$ whose center has order $$2$$, and whose commutator subgroup has order $$24$$. Then, any degree of an irreducible representation must divide $$24 = 48/2$$, and its square cannot be greater than $$24$$. This leads to the possibilities $$1,2,3,4$$ for degrees of irreducible representations. Further, there are exactly two irreducible representations of degree one, since the abelianization has order two. Thus, the remaining degrees are $$2,3,4$$. Suppose there are $$a_d$$ representations of degree $$d$$. We get:

$$4a_2 + 9a_3 + 16a_4 = 46$$.

This has two solutions: $$(a_2,a_3,a_4) = (7,2,0)$$ and $$(a_2,a_3,a_4) = (3,2,1)$$.

If we know the number of conjugacy classes in $$G$$, we can determine which of these cases occur.

Notice how we made so many deductions based on very limited information about the group.

Making progress with filling the character table
If a substantial portion of the character table has been filled in, we can use the orthogonality relations as well as facts about rationality and realization. Some of the tools are described here.

Using a linear representation we know that is not irreducible
Given a linear representation $$\rho$$ with character $$\chi$$, we can use the orthogonal projection formula to find out how many times each of the irreducible characters we already have occurs in $$\chi$$. Subtracting these off gives a character that is a sum of irreducible characters all of which are unknown at present. Next, we can calculate the norm of the (modified) character $$\chi$$, i.e., we compute:

$$\frac{1}{|G|} \sum_{g \in G} \left| \chi(g) \right|^2 = \sum_c |c| \left| \chi(g) \right|^2$$.

The first sum is over all group elements and the second sum is over conjugacy classes. Computing this sum requires us to known the sizes of the conjugacy classes.

This number is the sum of squares of the multiplicity of each of $$\chi$$'s irreducible constituents occurs in it. In other words, if we have:

$$\chi = a_1\psi_1 + \dots + a_r \psi_r$$

where the $$\psi_j$$ are irreducible, then the above norm is:

$$\sum_{i=1}^r a_i^2$$.

Again, the $$a_i$$ are integers, so we can usually determine what kind of irreducible constituents $$\chi$$ has and to what multiplicity. If $$\chi$$ itself is irreducible we are done. Otherwise, we might be able to use the above information to guess the irreducible constituents.

Using the orthogonality relations
If any one column is complete (e.g., the left-most column, which records the degrees) and another column has only one entry missing, we can use the column orthogonality theorem to deduce the missing entry. Similarly, if any one row is complete (which is always the case -- we have the principal character giving the first row) and another row has only one entry missing, we can fill in that entry. (For applying row orthogonality we need to know the sizes of the conjugacy classes).

More generally, if we have filled $$k$$ rows, we can fill in $$k$$ missing entries from any other row. We can often fill in $$k+1$$ missing entries, because we also have the condition of norm one. The analogous statement holds for columns.

Using the kernel and center to determine values on other conjugacy classes using values on a few
If $$g,h$$ are in distinct conjugacy classes, and the ratio $$g^{-1}h$$ is in the center, and the character value at that central element is $$d\lambda$$ where $$d$$ is the degree of the representation, then $$\chi(h) = \chi(g)\lambda$$.

Also, any element $$g$$ for which $$\chi(g)$$ equals the degree must be in the kernel, and any element $$g$$ for which the ratio of $$\chi(g)$$ and the degree is on the unit circle must map to a scalar matrix. Thus, we get the more general version:

If $$\chi$$ is of degree $$d$$ and $$\chi(g^{-1}h)/d$$ is a root of unity $$\lambda$$, then $$\chi(h) = \chi(g)\lambda$$.

Rationality considerations
Characters are cyclotomic integers, i.e., the values they take are sums of roots of unity. For groups of particular kinds, the rots of unity that we're allowed to consider are severely restricted.

For instance, an ambivalent group is a group for which every element is conjugate to its inverse. In an ambivalent group, every character is real-valued. A rational group is a group for which any two elements that generate the same cyclic subgroup are conjugate. In a rational group, every character is rational-valued, and thus, integer-valued. Examples of rational groups include symmetric groups, dihedral group:D8, and the quaternion group.

More generally, by a theorem of Brauer, every representation of a finite group can be realized over the cyclotomic field of $$d^{th}$$ roots of unity where $$d$$ is the exponent of $$G$$.

These rationality considerations place constraints on character values that we may be able to use to pinpoint them.

Finding the character table of a non-abelian group of order eight
Suppose $$G$$ is a non-abelian group of order eight. By observations made above, $$G$$ has center equal to its commutator subgroup, both having order two. It has four irreducible representations of degree one, and one irreducible representation of degree two. It's also easy to see the following:


 * Every element outside the center has a conjugacy class of size two, which is also a coset of the center.
 * The quotient $$G/Z(G)$$ is a Klein four-group (because the quotient by the center cannot be cyclic, as cyclic over central implies abelian).

Suppose $$e$$ is the identity element of $$G$$, $$z$$ is the central non-identity element, and the three conjugacy classes have elements $$a,az$$, $$b,bz$$, $$c,cz$$ respectively. There are three normal subgroups of order four, namely $$\{ e,z,a,az \}$$, $$\{ e,z,b,bz \}$$, $$\{ e,z,c,cz \}$$. The five conjugacy classes are $$\{ e \}, \{ z \}, \{ a,az \}, \{ b,bz \}, \{c, cz\}$$. This allows us to fill in the four one-dimensional representations: the trivial representation, and the three sign representations with the three normal subgroups as kernels.

There are many ways of filling in the final representation. One is purely using the column orthogonality relations, since we know that the entry in the identity column is $$2$$. Another is using the fact that since the center is not in the kernel of the two-dimensional representation, the non-identity central element must go to $$-2$$. Once we have this, we know that $$a$$ and $$az$$ are negatives of each other, so the character values at $$a$$ and $$az$$ are negative of each other. On the other hand, they are conjugate, so the character values at both elements are the same. This forces the character value to be zero. Similar reasoning shows that the character value at all the non-central conjugacy classes is zero.

Note that there are two possibilities for $$G$$ up to isomorphism: dihedral group:D8 and quaternion group, and the structure of these groups is different in many respects. However, the above computation shows that the two groups have the same character table.