Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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 $M$ be a $\mathbb{C}[G]$-module of the form $M=N\times N$, where $N$ is simple. How to conclude that $M$ has infinitly many direct sum decompositions into two copies of $N$ ?

This is what I have for now:

By Wedderburn's Theorem, one can assume that $M$ is a $M_{n \times n}(\mathbb{C})$-module. Since the only simple $M_{n \times n}(\mathbb{C})$-submodule is $\mathbb{C}^n$ up to isomorphism (because submodules need to be $M_{n \times n}(\mathbb{C})$-invariant, and the action of $M_{n \times n}(\mathbb{C})$ on the basis is by permutation). So we need to show $\mathbb{C}^n \times \mathbb{C}^n$ has infinitely many submodules.

share|cite|improve this question

Your Answer


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

Browse other questions tagged or ask your own question.