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Let $A$ be a finite dimensional semisimple algebra over $\mathbb{C}$ and $M$ is a finitely generated A-module. Prove that $M$ has only finitely many submodules iff $M$ is a direct sum of pairwise non-isomorphic simple modules.

Any hints for how to solve it?

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    @jspecter, $\mathbb{C}^2$ has infinitely many subspaces!2012-05-07
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    @CC_Azusa, use Artin-Wedderburn to reduce to the case where $A=M_n(\mathbb{C})$, and use the fact that any simple $A$-module is isomorphic to $\mathbb{C}^n$ with the natural action of $M_n(\mathbb{C})$. Show that $\mathbb{C}^n\oplus \mathbb{C}^n$ has infinitely many submodules.2012-05-07
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    @Bruno Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-10

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