Sorry for any mistakes I make here, this is my first post here. I have a group $G$ which has an abelian subgroup $A This is exercise 2.17 from Isaacs's Character Theory of Finite Groups, page 31. The hint is to show that $\chi$ vanishes on $G-A$.
Condition for abelian subgroup to be normal
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0I always forget the tricks for character theory. I don't know how much of help this is, and I'll try to back it up with a page reference, but I can almost guarantee you this is in the "Normal Subgroups" chapter of Isaac's *Character Theory*. Have you tried looking there? – 2011-10-28
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0Where is this from? Also, as stated, it is trivial, since the trivial subgroup is an abelian normal subgroup of any group. Do you mean that A must be normal, or that G must have an abelian normal subgroup of the same order as A? – 2011-10-28
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1I was going to post an answer that, at the end, used the Chermak-Delgado measure (See section 1G in Isaacs's *FGT*), but Nicky's comment that this is an exercise in the beginning of Isaacs's Character Theory book shows this is definitely not the right approach. – 2011-10-28
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0@SteveD: I wouldn't mind seeing it. I think if characters are mentioned, then characters should be used in the official answer, but alternatives might teach us a thing or two as well. – 2011-10-28
2 Answers
This is exercise (2.17) of Isaacs' Character Theory of Finite Groups. With Lemma (2.29) of that book you can see that $\chi$ vanishes outside $A$. Now look at the subgroup $N = \langle g \in G : \chi(g) \neq 0 \rangle$. Obviously $N \subseteq A$ hence $N$ is abelian. This subgroup is normal (conjugation does not change the character value) and non-trivial (otherwise the irreducible $\chi$ would vanish outside the identity element which is nonsense).
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0It is not immediate that $N$ is a subgroup. – 2011-10-28
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2@SteveD: It is immediate, but only because he has N be the subgroup generated by those g. All the generators lie inside A, so N lies inside A. – 2011-10-28
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0Ahh, yes, OK, I thought he was just talking about the set of such $g$. – 2011-10-28
Here is the approach I mentioned in the comments above:
We have an abelian subgroup $A Otherwise $Z(\chi)=\lbrace 1\rbrace$, and thus by Corollary 2.30 in Isaacs, we have $[G:A]^2 < |G|$, or equivalently, $|A|^2>|G|$. Now if we let $m(\cdot)$ be the Chermak-Delgado measure on $G$, then $m(A)\ge |A|^2>|G|$, and so if $M$ is the Chermak-Delgado subgroup, $m(M)>|G|$. Since $m(\lbrace1\rbrace)=|G|$, we must have $M$ is non-trivial.