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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.

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