Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Some of the applications of character theory are the proofs of Burnside $p^aq^b$ theorem, , Frobenius theorem and factorization of the group determinant (the problem which led Frobenius to character theory).

I would like to know what are some of the other problems which has been solved by the application of character theory especially from number theory and group theory.

share|improve this question
    
I guess many things proved using Dirichlet L-functions could be called an application of character theory. –  Alex J Best May 5 '13 at 16:01
    
Which Frobenius theorem are you talking about - Frobenius kernels or $x^n-1$? –  Alexander Gruber May 5 '13 at 16:05
    
Well,I had Frobenius kernels in mind but I think both can solved using character theory. –  Mohan May 5 '13 at 16:13
    
Big-list tag... community wiki? –  Stahl May 5 '13 at 19:05
add comment

5 Answers

@Alexander Gruber - A celebrated conjecture of Dedekind asserts that for any finite algebraic extension $K$ of $\mathbb{Q}$, the zeta function $\zeta_K(s)$ is divisible by the Riemann zeta function $\zeta(s)$. That is, the quotient $\zeta_K (s)/\zeta(s)$ is entire. More generally, Dedekind conjectures that if $L$ is a finite extension of $K$, then $\zeta_L(s)/\zeta_K(s)$ should be entire. This conjecture is still open, I believe.
By the work of Aramata and Brauer the conjecture is known to be true if $L/K$ is Galois. If $L$ is contained in a solvable extension of $K$, then Uchida and van der Waall have independently proved Dedekind’s conjecture. Proofs rely on properties of M-groups and hence this is a very nice application of character theory to algebraic number theory!

share|improve this answer
add comment

Character theory provides a better language to talk about certain group theoretic problems. Here is an example.

Definition. Let $G$ be a finite group.

  1. If $M\unlhd H \leqslant G$ and $H/M$ is cyclic, we call $(H,M)$ a pair.
  2. For $g\in G$ and $H\leqslant G$, write $$F_H(g)=\{[g,h]:h\in H\cap H^{g^{-1}}\}$$ If $(H,M)$ is a pair and $F_H(g)\not\subseteq M$ for all $g\in G$, we say that $(H,M)$ is a good pair.
  3. If $(H,M)$ and $(K,L)$ are good pairs, we say that they are related in $G$ if there exists a $g\in G$ for which $H^g\cap L=K \cap M^g$. It can be proven that being related in $G$ is an equivalence relation on the set of good pairs in $G$, and we denote by $m_G$ the number of equivalence classes under this relation.
  4. Denote by $n_G$ the number of equivalence classes of elements of $G$ under the relation $x\sim y \Leftrightarrow \langle x \rangle \text{ is conjugate to } \langle y \rangle\text{ in }G$.
  5. Let $\mathcal{M}$ be the class of finite groups with the property that $m_G=n_G$.

Goodness, that's a long definition. $\mathcal{M}$ is clearly a very difficult class of groups to study, impossible to understand. All the pieces fit together in some strange way, but it isn't quite clear what it all means.

Consider the following alternative definition.

Definition. A character of a finite group $G$ is monomial if it is induced from a linear character of a subgroup of $G$. We say $G$ is in the class $\mathcal{M}$ if every irreducible character of $G$ is monomial.

The groups in both these definitions are called $\mathcal{M}$-groups, with proof of equivalence given yonder. We know a lot of things about them: $\mathcal{M}$-groups are solvable, supersolvable groups are $\mathcal{M}$-groups, normal subgroups of $\mathcal{M}$-groups and their quotients are both $\mathcal{M}$-groups, every solvable group can be embedded in an $\mathcal{M}$-group.

It is not at all obvious that these definitions are equivalent. The first definition was introduced $55$ years after the second, motivated by informal questions posed by several people who studied $\mathcal{M}$-groups using pairs. The purpose of this post is not to devalue the study of pairs in $\mathcal{M}$-groups, but to emphasize that character theory provides an easier language in which to study some problems in group theory, even when it is not logically necessary as in the case of Burnside's and Frobenius' theorems.

share|improve this answer
    
This doesn't feel much like an "application" of character theory to me. I do think M-groups are interesting, but the reason they are interesting is because of the relation to character theory! –  S123 May 5 '13 at 16:56
    
@Steve I concede this is true! My point was to illustrate a situation in which character theory is not absolutely logically necessary to prove something in group theory (such as Burnside's/Frobenius' theorems), but is still a better tool for doing so. I plan to write another post containing more applications in the traditional sense. –  Alexander Gruber May 5 '13 at 16:59
add comment

Representations of finite groups can be used to prove Hurwitz's 1,2,4,8 theorem. See http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/hurwitzrepnthy.pdf.

share|improve this answer
add comment

The Feit–Thompson theorem, or Odd Order Theorem, states that every finite group of odd order is solvable. It was proved by Walter Feit and John Thompson (1963). Its proof ($2^8-1$ pages long) relies heavily on character theory.

share|improve this answer
add comment

One of the prominent applications of representation theory of the symmetric group is to the study of random walks on groups. Look for instance Chapter 11 in 'Representation Theory of Finite Groups' by Benjamin Steinberg.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.