Being a subgroup: proof by character theory

Question

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 1

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 2

Let me re-phrase my remark.

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

Answer 3

A result of a different nature is something called the Brauer trick, due (of course) to R. Brauer. It is a way to show that two (not necessarily normal) proper subgroups $A$ and $B$ of a finite group $G$ together generate a proper subgroup of $G,$ using a character-theoretic argument.

The idea is to take a complex irreducible (non-trivial) character $\chi$ of $G$, and try to show that $$\langle {\rm Res}_{A}^{G}(\chi),1\rangle + \langle {\rm Res}_{B}^{G}(\chi),1\rangle > \langle {\rm Res}_{A \cap B}^{G}(\chi),1\rangle .$$

If that inequality holds, then in the underlying $\mathbb{C}G$-module, $V$ say, associated with $\chi,$ we see $C_{V}(A)$ and $C_{V}(B)$ are both contained in $C_{V}(A \cap B),$ so by elementary linear algebra $$C_{V}(A) \cap C_{V}(B) \neq \{0\}.$$ In that case $\langle A,B \rangle$ fixes a non-zero vector $v$, so since $\chi$ is irreducible, the subgroup $\langle A,B \rangle$ is a proper subgroup of $G$.

Of course, here, there is no issue of whether $\langle A,B \rangle$ is a subgroup, but (when applicable) the character theory demonstrates that this subgroup is proper.

Answer 4

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.