Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a finite group and $N$ be a normal subgoup of G. Suppose that $\chi \in Irr(G)$. If $\theta , \lambda \in Irr(N)$, such that $[\chi_{N}, \theta]>0$ , $[\chi_{N}, \lambda]>0$, is it true that $\theta(1)=\lambda(1)$?

On the other hand, irreducible constituents of $\chi_{N}$ are uniqe?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Yes, it is true ( all irreducible constituents of the restriction of an irreducible character to a normal subgroup all have equal degree). This is part of Clifford's theorem. It actually applies not just to complex irreducible characters or representations, but to irreducible representations over any field. In the case of complex characters, the uniqueness of the consituents of the restricted character follows because of the fact that the irreducible characters of $N$ form an orthonormal basis for the space of class functions of $N$ with respect to the usual inner product of class functions. For representations over other fields, the Jordan Holder theorem can also be used.

Later edit: Clifford's theorem can be found in many texts: the module version states that if $S$ is a simple $FG$-module where $G$ is a finite group and $F$ is a field, and we have a normal subgroup $N$ of $G$, then ${\rm Res}^{G}_{N}(S)$ is semisimple, and all simple$FN$-summands are conjugate under the action of $G$ (so, in particular, all have the same $F$-dimension) and al occur with the same multiplicity as summands of the restricted module.

share|improve this answer

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.