# Prove that every finite group of order larger than 2 has more than two irreducible complex representations. [closed]

prove that every finite group of order larger than 2 has more than two irreducible complex representations. could anyone give me a hint on how to solve this please?

## closed as off-topic by José Carlos Santos, Eevee Trainer, mrtaurho, Thomas Shelby, Paul FrostMar 9 at 9:24

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• Are you aware that the number of irreducible complex representations of a finite group is equal to the number of conjugacy classes? – Ethan Alwaise Mar 9 at 7:15
• yes I know this @EthanAlwaise – Secretly Mar 9 at 8:21

Let $$G$$ be a finite group of order $$n$$ with conjugacy classes $$C_1,\ldots,C_k$$. Then $$\vert C_i \vert$$, the order of the conjugacy class $$C_i$$, divides $$n$$. The conjugacy classes also partition $$G$$, so we have $$\vert C_1 \vert + \cdots + \vert C_k \vert = n.$$ Since the number of irreducible complex representations of $$G$$ is equal to $$k$$, if $$G$$ has less than three irreducible complex representations, then $$k \leq 2$$. So either $$k = 1$$ which implies $$G$$ is trivial, or $$k = 2$$ and we get the equation $$\vert C_1 \vert + \vert C_2 \vert = n.$$ However, one of the conjugacy classes, WLOG $$C_1$$, is the class of the identity and thus has size $$1$$. So we have the equation $$1 + \vert C_2 \vert = n,$$ hence $$\vert C_2 \vert = n - 1$$. The only $$n$$ for which $$n - 1$$ divides $$n$$ is $$n = 2$$, so $$G$$ is the cyclic group of order two.