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Assume that $G$ is an infinite group with center of finite index. How do we prove that the degree of a finite dimensional irreducible representation of $G$ over $\mathbb{C}$ divides the index of the center ? or is it false ? what is known about this question ?

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Let $G$ be the group with center $Z_G$. By Schur's lemma $Z_G$ acts by scalar matrices. Thus we have an irreducible projective representation of the quotient $G/Z_G$ which is a finite group. Now, your question follows from the following fact:

Proposition Let $\Gamma$ be a finite group and let $V$ be a irreducible finite-dimensional projective representation of $\Gamma$. Then $\dim(V)$ divides the order $|\Gamma|$.

Proof It is known (and I think goes back to Schur) that any projective representation $V$ of $\Gamma$ lifts to a representation of some finite central cover $\widetilde\Gamma$ of $\Gamma$. Now $\dim(V)$ divides the order of $|\widetilde\Gamma/Z_{\widetilde\Gamma}|$ as proved in e.g. Section 6.4 of Serre's book. But $\Gamma$ maps surjectively onto $\widetilde\Gamma/Z_{\widetilde\Gamma}$ so $\dim(V)$ also divides $|\Gamma|$.

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    $\begingroup$ Maybe (?) an equivalent form of the answer is that because $Z_G\rightarrow G/(G, G)$ has finite kernel, after twisting by a character the representation becomes trivial on a finite index subgroup of $Z_G$. Therefore the assertion follows from the same proposition in Serre’s book :p $\endgroup$ Commented Jan 22 at 9:27
  • $\begingroup$ @Cheng-ChiangTsai that's a very nice argument! $\endgroup$ Commented Jan 22 at 9:33
  • $\begingroup$ @KentaSuzuki I remarked that u can also use the preimage of the image of $\Gamma\subset PGL(V)$ under the surjective morphism $SL(V)\to PGL(V)$. $\endgroup$ Commented Jan 22 at 17:49
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    $\begingroup$ @KentaSuzuki Yay! I still feel like the related elements appeared in your talk two weeks ago :p $\endgroup$ Commented Jan 23 at 2:35

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