## Relation Between the Kernels of Characters and Normal Subgroups

**Point of post: **In this post we derive some fundamental relationships between the kernels of the characters of a finite group and the set of all normal subgroups of . In particular, we prove that every normal subgroup of has the realization as the kernel of some character, and thus by previous theorem that every normal subgroup of has a realization as the intersection of certain ‘s.

*Motivation*

We saw in our last post that there is a particularly fruitful way to produce normal subgroups ofa finite group . Namely, one can look at the kernels of the characters of , which as was shown amounts (in the sense that all the necessary information is contained in) to finding the kernels of the irreducible characters of . What we shall now show is that in fact, finding the kernels of the irreducible characters of a group enables us to list off precisely what normal subgroups has. In particular, we will see that every normal subgroup of has a realization as the kernel of a certain character. We shall do this by constructing a certain character on the quotient group of for which, when thought of a character of , will have kernel equal to the normal subgroup.

*Natural Character on the Quotient Group*

Let be a finite group and . Consider then the quotient group and its associated group algebra . We define then by

this is evidently a representation since

Moreover, it’s evident and moreover if one has that and so in particular and so or . It follows that . It follows then that if is the associated character of that . We sum this up in the following theorem:

**Theorem: ***Let be a finite group and . Then, there exists some character of for which .*

Combining this and our previous results about the kernels of general characters we can conclude the following:

**Corollary: ***Let be a finite group. Then,*

*Relationship Between the Characters of a Group and the Cardinalities of its Normal Subgroups*

We can do more than just find the set of all normal subgroups of by inspecting the kernels of its irreducible characters, we can easily apply previous results about characters to ascertain the orders of all such subgroups. While this is mildly redundant it does give us the interesting result that a group is simple if and only if the kernel of every non-trivial character is trivial. Namely

**Theorem: ***Let be a finite group and . Then, if are the conjugacy classes such that then *

**Proof: **By definition one has that

but by the second orthogonality relation one has that

and so

The conclusion follows.

**Corollary: ***Let be a finite group. Then, is simple if and only if for every non-trivial character .*

**References:**

1. Isaacs, I. Martin. *Character Theory of Finite Groups*. New York: Academic, 1976. Print.

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