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 $G-A$.
 A: 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).
A: 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.
