Abstract Nonsense

Crushing one theorem at a time

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 G and the set of all normal subgroups of G. In particular, we prove that every normal subgroup of G has the realization as the kernel of some character, and thus by previous theorem that every normal subgroup of G has a realization as the intersection of certain N^{(\alpha)}‘s.

Motivation

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

Natural Character on the Quotient Group

Let G be a finite group and N\unlhd G. Consider then the quotient group G/N and its associated group algebra \mathcal{A}\left(G/N\right). We define then \rho:G\to\mathcal{A}\left(G/N\right) by

\left(\rho_g(f)\right)(hN)=f\left(hNgN\right)=f\left(hgN\right)

 

this is evidently a representation since

\left(\rho_{gh}(f)\right)(k)=f\left(kghN\right)=\left(\rho_g\left(\rho_h(f)\right)\right)\left(k\right)

 

Moreover, it’s evident N\subseteq\ker\rho and moreover if g\in\ker\rho one has that \rho_g(\delta_{N})=\delta_{N} and so in particular 1=\delta_N(N)=\delta_N(gN) and so N=gN or g\in N. It follows that g\in N. It follows then that if \chi_\rho is the associated character of \rho that \ker\chi=N. We sum this up in the following theorem:

Theorem: Let G be a finite group and N\unlhd G. Then, there exists some character \chi of G for which N=\ker\chi.

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

Corollary: Let G be a finite group. Then,

\displaystyle \left\{\text{Normal Subgroups of }G\right\}=\left\{\bigcap_{\alpha\in\Omega}\ker\chi^{(\alpha)}:\Omega\subseteq\widehat{G}\right\}

 

 

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 G 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 G is simple if and only if the kernel of every non-trivial character is trivial. Namely

Theorem: Let G be a finite group and N\unlhd G. Then, if \mathcal{C}(g_1),\cdots,\mathcal{C}(g_m) are the conjugacy classes such that \displaystyle N=\biguplus_{j=1}^{m}\mathcal{C}(g_j) then

\displaystyle |N|=|G|\sum_{j=1}^{m}\left(\sum_{\alpha\in\widehat{G}}\left|\chi^{(\alpha)}(g_j)\right|^2\right)^{-1}

Proof: By definition one has that

\displaystyle |N|=\sum_{j=1}^{m}\#\left(\mathcal{C}(g_j)\right)=\sum_{g\in G}\frac{|G|}{\bold{C}_G(g_j)}

 

but by the second orthogonality relation one has that

\displaystyle \bold{C}_G(g_j)=\sum_{\alpha\in\widehat{G}}\left|\chi^{(\alpha)}(g_j)\right|^2

 

and so

\displaystyle |N|=\sum_{j=1}^{m}|G|\left(\sum_{\alpha\in\widehat{G}}\left|\chi^{(\alpha)}(g_j)\right|^2\right)^{-1}=|G|\sum_{j=1}^{m}\left(\sum_{\alpha\in\widehat{G}}\left|\chi^{(\alpha)}(g_j)\right|^2\right)^{-1}

 

The conclusion follows. \blacksquare

Corollary: Let G be a finite group. Then, G is simple if and only if \ker\chi=\{e\} for every non-trivial character \chi.

References:

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

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March 7, 2011 - Posted by | Algebra, Group Theory, Representation Theory | , , , , ,

4 Comments »

  1. […] theorem that for every . Thus, for any one has that or that . It follows then that . But, recalling that every normal subgroup is the realization of some intersection of the ‘s, and so in […]

    Pingback by Representation Theory: The Center of a Character « Abstract Nonsense | March 8, 2011 | Reply

  2. […] theorem that for every . Thus, for any one has that or that . It follows then that . But, recalling that every normal subgroup is the realization of some intersection of the ‘s, and so in […]

    Pingback by Representation Theory: The Center of a Character (Pt. II) « Abstract Nonsense | March 9, 2011 | Reply

  3. […] Proof: This follows from our work with the kernels of characters. […]

    Pingback by Representation Theory: Character Tables « Abstract Nonsense | March 21, 2011 | Reply

  4. […] CT 5 and CT 6 which are really just statements about the kernel and center of a character respectively we may conclude that the set of all normal subgroups of is […]

    Pingback by Representation Theory: Character Table of S_3 Without Finding Irreducible Characters « Abstract Nonsense | March 22, 2011 | Reply


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