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 Frost Mar 9 at 9:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, mrtaurho, Thomas Shelby, Paul Frost
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.