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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?

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

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

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