Being a subgroup: proof by character theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:30:19Z http://mathoverflow.net/feeds/question/80127 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80127/being-a-subgroup-proof-by-character-theory Being a subgroup: proof by character theory Denis Serre 2011-11-05T11:10:22Z 2012-12-13T14:13:05Z <p>Let me first cite a theorem due to Frobenius:</p> <blockquote> <p>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 $G$.</p> </blockquote> <p>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!</p> <blockquote> <p>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?</p> </blockquote> <p><strong>Edit</strong>: 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</p> http://mathoverflow.net/questions/80127/being-a-subgroup-proof-by-character-theory/80131#80131 Answer by John mac for Being a subgroup: proof by character theory John mac 2011-11-05T13:21:38Z 2011-11-05T13:21:38Z <p>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. </p> http://mathoverflow.net/questions/80127/being-a-subgroup-proof-by-character-theory/80156#80156 Answer by DavidLHarden for Being a subgroup: proof by character theory DavidLHarden 2011-11-05T21:27:47Z 2011-11-05T21:27:47Z <p>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): </p> <p>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 $G$. </p> <p>For the proof, see </p> <p>Nobuo Iiyori and Hiroyoshi Yamaki, On a conjecture of Frobenius, Bulletin of the American Mathematical Society (New Series) 25 (1991), no. 2, 413-416 . </p> <p>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...</p> http://mathoverflow.net/questions/80127/being-a-subgroup-proof-by-character-theory/80972#80972 Answer by John McKay for Being a subgroup: proof by character theory John McKay 2011-11-15T10:52:14Z 2011-11-15T10:52:14Z <p>Let me re-phrase my remark.</p> <p>Give sufficient conditions for a character to be a permutation character. </p>