# Changes

## Determining the character table of a finite group

, 21:27, 12 April 2009
no edit summary
These rationality considerations place constraints on character values that we may be able to use to pinpoint them.

==Some worked out examples==

===Finding the character table of a non-abelian group of order eight===

Suppose [itex]G[/itex] is a non-abelian group of order eight. By observations made [[#some examples|above]], [itex]G[/itex] 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 [itex]G/Z(G)[/itex] is a [[Klein four-group]] (because the quotient by the center cannot be cyclic, as [[cyclic over central implies abelian]]).

Suppose [itex]e[/itex] is the identity element of [itex]G[/itex], [itex]z[/itex] is the central non-identity element, and the three conjugacy classes have elements [itex]a,az[/itex], [itex]b,bz[/itex], [itex]c,cz[/itex] respectively. There are three normal subgroups of order four, namely [itex]\{ e,z,a,az \}[/itex], [itex]\{ e,z,b,bz \}[/itex], [itex]\{ e,z,c,cz \}[/itex]. The five conjugacy classes are [itex]\{ e \}, \{ z \}, \{ a,az \}, \{ b,bz \}, \{c, cz\}[/itex]. 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 [itex]2[/itex]. 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 [itex]-2[/itex]. Once we have this, we know that [itex]a[/itex] and [itex]az[/itex] are negatives of each other, so the character values at [itex]a[/itex] and [itex]az[/itex] 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 [itex]G[/itex] 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.