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Let $A$ be a finite-dimensional simple algebra over $\mathbb{C}$ of dimension $n$. By Wedderburn's theorem, we have that $A$ is isomorphic to a matrix ring $M_r(\mathbb{C})$, which is of dimension $r^2$ over $\mathbb{C}$, which would imply that $n$ the dimension of $A$ over $\mathbb{C}$ is necessarily square.

This is false, however, as there exist irreducible representations of group algebras $\mathbb{C}G$ with non-square (and in fact square-free) dimension.

Where is my confusion here?

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    $\begingroup$ $\mathbb{C}G$ is semisimple, not simple (unless $G$ is the trivial group). $\endgroup$ – darij grinberg Feb 28 '15 at 17:51
  • $\begingroup$ Yes, but irreducible representations of $G$ are simple. $\endgroup$ – wedderburndimension Feb 28 '15 at 17:53
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    $\begingroup$ Oh, I see what you mean. But irreducible representations of $G$ are $\mathbb CG$-modules, not $\mathbb C$-algebras. That's a different beast. There is no Artin-Wedderburn theorem for modules. $\endgroup$ – darij grinberg Feb 28 '15 at 17:56
  • $\begingroup$ I see, thanks! The simple submodules of $\mathbb{C}G$ are generally not themselves simple rings. $\endgroup$ – wedderburndimension Feb 28 '15 at 18:06
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The (unique) left simple modulo of the algebra $M_r(\Bbb{C})$ is the column space $\Bbb{C}^r$. In other words the simple component of dimension $r^2$ has a simple module of dimension $r$.

Observe that in case of a finite group $G$ of order $n$ and dimensions of simple representations $d_1,d_2,\ldots,d_k$ we have the equation $$ n=d_1^2+d_2^2+\cdots+d_k^2. $$ It all fits magically together :-)

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