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Suppose the order of $G$ is $n < \infty$. Suppose $R$ is an irreducible representation of $G$, what can we conclude about its dimension with relation to the order of $G$?

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    $\begingroup$ The sum of the squares of the dimensions of the irreducible representations of $G$ over $\Bbb C$ is just $\#G$, so, crudely, any representation has dimension $\leq \sqrt{\#G}$. It's also true that the dimension of any such representation divides $\#G$. (Note that these things need not be true for representations over nonclosed fields, e.g., $\Bbb Q$.) $\endgroup$ Apr 18 '17 at 16:19
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Let me assume you are working with complex vector spaces. Sum of squares of degrees of all the irreducible representations equals the order of the group.

Another theorem is any of these degrees divide the order of the group, in fact they also divide the index of the centre of the group. (See Serre's book on Representation of Finite Groups).

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