# Condition for abelian subgroup to be normal

Sorry for any mistakes I make here, this is my first post here. I have a group $G$ which has an abelian subgroup $A<G$. I also know there is a irreducible character $\chi$ with the degree of $\chi$ equal to the index of $A$ in $G$. This implies $G$ has a non-identity abelian NORMAL subgroup? How?

This is exercise 2.17 from Isaacs's Character Theory of Finite Groups, page 31.

The hint is to show that $\chi$ vanishes on $G-A$.

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I 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? – Alex Youcis Oct 28 '11 at 7:02
Where 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? – Tobias Kildetoft Oct 28 '11 at 8:38
I 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. – user641 Oct 28 '11 at 15:44
@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. – Jack Schmidt Oct 28 '11 at 17:12

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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It is not immediate that $N$ is a subgroup. – user641 Oct 28 '11 at 14:15
@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. – Jack Schmidt Oct 28 '11 at 15:46
Ahh, yes, OK, I thought he was just talking about the set of such $g$. – user641 Oct 28 '11 at 18:31
thanks so much for your help Nicky. And others. – Jason Rhodes Oct 30 '11 at 2:10

Here is the approach I mentioned in the comments above:

We have an abelian subgroup $A<G$, and we also have an irreducible character $\chi$ such that $\chi(1)=[G:A]$. From Lemma 2.29 in Isaacs, we see that $\chi$ must be zero outside of $A$. Now if the center of the character $Z(\chi)$ is non-trivial, then it is a normal subgroup of $G$, contained in $A$, so we are done.

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.

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