As alluded to at the end of the previous article, we shall consider the case where *K* is algebraically closed, i.e. every polynomial with coefficients in *K* factors as a product of linear polynomials. E.g. *K* = **C** is a common choice.

Having assumed this, we see that any division ring *D* ⊇ *K* which is finite-dimensional as a *K*-vector space must be *K* itself; indeed, if *x* ∈ *D*, then *K*(*x*) is a commutative division ring (i.e. a field) over *K* which is of finite dimension, so it must be *K* itself, and *x* ∈ *K*. Thus,

and for any simple *K*[*G*]-module *V*, we have End_{K[G]}(*V*) = *K*. This also gives:

,

where *V*_{i} runs through all simple *K*[*G*]-modules up to isomorphism.

In particular, this mean for any simple *K*[*G*]-module *V*. Thus, their characters actually form an *orthonormal* set over *K*. The next theorem we want to show is:

Theorem. The set of , where V runs through all simple K[G]-modules, forms an orthonormal basis of the space of class functions.

**Proof**

We need to show the characters span the space of class functions. Let *f* : *G* → *K* be any class function; subtracting by a linear combination of characters, we may assume *f* is orthogonal to all Let

Since *f* is a class function, we get for any *g* in *G*. Thus *α *commutes with every element of *K*[*G*]. If *V* is a simple *K*[*G*]-module, let *m* : *V* → *V* be the multiplication-by-*α* map; since *α* commutes with every element of *K*[*G*], *m* is a *K*[*G*]-linear map. From above, we know that End_{K[G]}(*V*) = *K*, so *m* is a scalar map represented by λ ∈ *K*.

To find this λ, we take the trace of *m* :

which is zero since *f* was picked to be orthogonal to all characters. Hence multiplication-by-*α* is the zero map for all simple *K*[*G*]-module *V*; since *K*[*G*] is semisimple, multiplication-by-*α* is also zero on it. But that map takes *e* to *α* so *α*=0. ♦

Summary. This is what we know so far:

- The number k of isomorphism classes of simple K[G]-modules is precisely the number of conjugancy classes of G (by the above theorem).
- Their dimensions satisfy
- Their characters form an orthonormal basis for the space of class functions G → K.

As runs through all characters of simple modules and *g* runs through conjugancy classes, we obtain a *k* × *k* tables for the values Here are the **character tables** for *S*_{4} and *S*_{5}:

The number [*n*] in square brackets refers to the number of elements in each conjugancy class. Let us check that and are orthogonal:

For the details of computations, refer to an earlier article. Since both character tables comprise of solely integers, one suspects that in fact the modules are defined over **Q**, i.e. that the characters of simple **Q**[*G*]-modules are already *orthonormal*. This is, in fact, true for any symmetric group *S _{n}* – but we will have to revisit this another day.

Finally, we end this article with an example of a simple *K*[*G*]-module *V* where End(*V*) is a non-commutative division ring.

## Division Ring Example

Let *K* = **R**, and consider the division ring of quaternions **H**. Let *G* = {±1, ±i, ±j, ±k} ⊂ **H** which is a non-abelian group of order 8.Let’s compute its character table. First note that *G* has 5 conjugancy classes: {+1}, {-1}, {±i}, {±j}, {±k}. Next:

- There’s the trivial representation, where all
*g*→ 1. - Let
*N*= {±1, ±i}, which is of index 2 in*G*and thus normal. The non-trivial representation of*G*/*N*≅*C*_{2}→**R*** then pulls back to a representation of*G*which is irreducible since it has dimension 1. - Do the same for {±1, ±j} and {±1, ±k}.

This gives us 4 irreducible representations of dimension 1. Let us compute the character table over **C**. Since the last character has dimension 2. The other entries are easy to compute:

So this is for **C**, but what about over **R**? Note from our description above that the 4 characters of dimension 1 are defined over **R**. Write

where the character τ of *W* is given by Is *W* simple?

Note that upon extension to **C**, we have where *V* is the simple **C**[*G*]-module with character χ = (2, -2, 0, 0, 0) above. Thus if *W* is not simple, we must have for some **R**[*G*]-module *V*, such that dim_{R}*V* = 2.

We shall prove that such a *V* does not exist; let ρ : *G* → GL(2, **R**) be the corresponding group homomorphism.

- We have ρ(+1) = I and ρ(-1) = -I (the latter because it has trace -2 from the character table).
- The other 6 elements all satisfy so and so ρ(
*g*) has eigenvalues ±√-1. From the character table, its trace is zero so the eigenvalues are +√-1 and -√-1. - Thus ρ(
*g*) has det = 1 and trace = 0, corresponding to the product and sum of its eigenvalues. - It remains to show that if
*A*,*B*are 2×2 real matrices with det = 1 and trace = 0, then the same cannot hold fo*AB*.

For that, it is easy to show *A* and *B* are of the form:

Upon expanding tr(*AB*) we obtain:

Over the reals, we must have *b* = *d* = 0, which is absurd. Thus, *V* does not exist so *W* is already simple as an **R**[*G*]-module. Hence the structure of the group ring **R**[*G*] is:

where *D* is a division ring of dimension 4 over **R**. [ In fact, it is the ring of quaternions. ]