Before giving the definition of the character table of a finite group we need some preparation. We defined an irreducible character: is called irreducible if is an irreducible representation. We also saw that equivalent representations have equal characters and we mentioned that the converse of this fact is also true, which we will prove it later (see Remark 2 in here). We proved that if has conjugacy classes, then the number of non-equivalent irreducible representations of is exactly (see the theorem in here). Thus the number of distinct irreducible characters of is exactly Finally, recall that the value of a character is constant on each conjugacy class (see part iii) of Remark 1 in here).

Let be the set of conjugacy classes of where we will choose to be the conjugacy class of the identity element of So Let be the set of distinct irreducible characters of We will choose to be the trivial character of i.e. is the character afforded by the trivial representation defined by for all So for all We are now ready to define the character table of

**Definition**. The **character table** of is the table (or matrix) defined by for all where is any element of

**Remark 1**. Each entry in the first row of is because, since by convention is the trivial character of we have The -th entry in the first column of is because by part i) of Remark 1 in here.

**Example**. Find the character table of

**Solution**. We know that is isomorphic to the dihedral group of order Thus we can write where we choose and to be cycles of length 2 and 3 respectively. We know from group theory, and it’s easy to see, that has conjugacy classes:

So has three non-equivalent irreducible representations affording distinct irreducible characters where is the trivial character. So our character table is In Example 2 in here we determined and We showed that in fact in general, has only two representations of degree one: the trivial representation and Let Then, the third representation has degree two and is defined by

for all and Let’s pick an element of each conjugacy class: and Of course, you may choose any element you like and that wouldn’t change the character table. Since is trivial, we have as we also mentioned in the above remark. Let’s now find the values of Well, we have Now, since is a cycle of length two, it is odd and hence its signature is Thus The cycle has length three and so it is even. Thus Finally, let’s find the values of We have Now, from and the fact that it is clear that and So, the character table of is

**Remark 2**. It is possible for two non-isomorphic groups to have the same character tables; e.g. and We will show this later. So character tables do not determine groups up to isomorphism.