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Denis Serre
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Being a subgroup: proof by character theory

Let me first cite a theorem due to Frobenius:

Let $G$ be a finite group, with $H$ a proper subgroup ($H\ne (1)$ and $G$). Suppose that for every $g\not\in H$, we have $H\cap gHg^{-1}=(1)$. Then $$N:=(1)\cup(G\setminus\bigcup_{g\in G}gHg^{-1})$$ is a normal subgroup of $G$.

The proof is fascinating. One never proves directly that $N$ is stable under the product and the inversion. Instead, one constructs a complex character $\chi$ over $G$, with the property that $\chi(g)=\chi(1)$ if and only if $g\in N$. This ensures (using the equality case in the triangle inequality) that the corresponding representation $\rho$ satisfies $\rho(g)=1$ if and only if $g\in N$. Hence $N=\ker \rho$ is a subgroup, a normal one!

Does anyone know an other example where a subset $S$ of a finite group $G$ is proven to be a subgroup (perhaps a normal one) by using character theory? Is there any analogous situation when $G$ is infinite, say locally compact or compact?

Edit: If the last argument, in the proof that $S$ is a subgroup, is that $S$ is the kernel of some character, then $S$ has to be normal. Therefore, an even more interesting question is whether there is some (family of) pairs $(G,T)$ where $T$ is a non-normal subgroup of $G$, and the fact that $T$ is a subgroup is proved by character theory. I should be happy to have an example, even if there is another, character-free, proof

Denis Serre
  • 52.3k
  • 10
  • 146
  • 300