Being a subgroup: proof by character theory - MathOverflow

Being a subgroup: proof by character theory

2011-11-05T11:10:22Z

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

Answer by John mac for Being a subgroup: proof by character theory

2011-11-05T13:21:38Z

More enticing is the question of determining the existence of a subgroup H (which need not be normal) of G from the character table of G.

Answer by DavidLHarden for Being a subgroup: proof by character theory

2011-11-05T21:27:47Z

There is another example of sorts, but it's not a good example since it appeals to a result known only as a consequence of the Classification of Finite Simple Groups (whose proof involves A LOT of character theory, instead of one more new technique, or one more variation on an old basic character-theoretic technique). It does, however, strictly generalize that theorem of Frobenius (by letting $n$ be the order of the Frobenius kernel):

Let $G$ be a finite group, and suppose $n$ is a positive integer dividing $|G|$. If the number of solutions in $G$ to $x^{n} = 1$ is exactly $n$, these solutions form a subgroup of $G$.

For the proof, see

Nobuo Iiyori and Hiroyoshi Yamaki, On a conjecture of Frobenius, Bulletin of the American Mathematical Society (New Series) 25 (1991), no. 2, 413-416 .

As with the theorem of Frobenius which this result generalizes, it is easy to prove this subset of $G$ contains the identity and is closed under taking inverses. So the only difficulty is in proving closure under composition...

Answer by John McKay for Being a subgroup: proof by character theory

2011-11-15T10:52:14Z

Let me re-phrase my remark.

Give sufficient conditions for a character to be a permutation character.