MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
up vote 6 down vote accepted

Yes, it is true ( the 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 all occur with the same multiplicity as summands of the restricted module.

share|cite|improve this answer

Your Answer


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.