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

  • $\begingroup$ Do you have any reference for this theorem ? $\endgroup$ Nov 5, 2011 at 12:22
  • $\begingroup$ @Paul. Shamely, ... my reference is the course about finite groups given by Jean-Pierre Serre in 78/79 at Sèvres, and posted recently on ArXiv (in French). See the chapter 6 about Frobenius subgroups. $\endgroup$ Nov 5, 2011 at 12:53
  • $\begingroup$ Geoff Robinson comments on the case of infinite groups underneath his answer to this question of mine: mathoverflow.net/questions/63142/…. Also, note that chapter 7 of Isaacs is all about generalisations of the basic ideas that go into Frobenius's proof (and of course the proof itself). $\endgroup$
    – Alex B.
    Nov 5, 2011 at 13:01
  • 2
    $\begingroup$ BTW, the arxiv link to Serre's notes is this: arxiv.org/abs/math/0503154 $\endgroup$ Nov 6, 2011 at 2:23
  • $\begingroup$ The referenced theorem is also mentioned and proved in Isaacs' "Character Theory of Finite Groups". It is Theorem 7.2. $\endgroup$ Dec 13, 2012 at 14:36

3 Answers 3


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...


Let me re-phrase my remark.

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

  • $\begingroup$ It looks like you have more than one account; another is under the name John mac. We can merge these into a single account if you like. $\endgroup$
    – Todd Trimble
    Sep 28, 2015 at 12:21

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.

  • $\begingroup$ Trivially, every group has a subgroup $H$, e.g. $H=G$. So I am not sure what you wanted to say. $\endgroup$
    – GH from MO
    Nov 5, 2011 at 15:40
  • $\begingroup$ Presumably, he meant a proper nontrivial subgroup? $\endgroup$
    – Will Sawin
    Nov 5, 2011 at 16:12
  • $\begingroup$ @Will: That would be equally trivial. The only groups that don't have proper nontrivial subgroups are the groups of prime order. Those groups are of course very, very easy to recognize by their character table. I think John mac though in his comment (it isn't really an answer) more about specific types of subgroups. Questions like "Does there exists a centralizer of an involution that has properties X and Y but not property Z" are very interesting and it would certainly be an achievement if one were able to provide a general answer in terms of the character table. $\endgroup$ Nov 5, 2011 at 17:53
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    $\begingroup$ Perhaps he means you are just given the character table of some group $G$, and you know nothing else about it: can you list (some of) the subgroups, and to what extent? I take his comment to be a generalization of the well-known fact that one can read off normal subgroups from the character table. $\endgroup$
    – Steve D
    Nov 5, 2011 at 18:10

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