The condition between $\chi(1)$ and $[G:H]$ which gives us a normal subgroup. $\textbf{The question is as follows:}$

Let $H \le G$ with $|G : H| = n$ and suppose that $\chi \in Char(G)$.
  $\rm (a)$ Show that $[\chi ; \chi] \ge \frac{[\chi_H; \chi_H]}{n}$.
  $\rm (b)$ Show that, if $H$ is Abelian and $\chi \in Irr(G)$ then $\chi(1) \le n$.
  $\rm (c)$ Show that, if the equality holds in part (b), then $H \vartriangleleft G$.

I can show the first part as follows:
$\rm (a)$  We have
$$|H| [\chi_H; \chi_H] = \sum_{h \in H}^{} |\chi(h)|^2 \le \sum_{g \in G}^{}|\chi(g)|^2 = |G|[\chi, \chi] $$
$\rm (b)$ For second part I can write
$$\chi(1)=\chi|_H(1)\le [\chi|_H,\chi|_G]\le [G:H][\chi,\chi]=[G:H]$$
But I am not sure if it is correct?
Can someone correct me please?
$\rm (c)$  For the third part I still have no idea so far!
Can someone help me to show this, please?
Thanks!
 A: Any reference here is to the book of Marty Isaacs, Character Theory of Finite Groups. Now for (b) you must be a bit more precise: if $H$ is abelian, and $\chi$ is an irreducible character of $G$, then $\chi_H=\sum_{\lambda \in Irr(H)}a_{\lambda}\lambda$, where the $\lambda$'s are linear and the integers $a_\lambda \geq 0$. Hence, $[\chi_H,\chi_H]=\sum_{\lambda \in Irr(H)}a_{\lambda}^2 \geq \sum_{\lambda \in Irr(H)}a_{\lambda}=\chi(1).$ By (a) and the fact that $\chi$ is irreducible so $[\chi,\chi]=1$, you get indeed $\chi(1) \leq [\chi_H,\chi_H] \leq |G:H|. \text{  } (*)$
For (c) we are going to use the Lemma (2.29) of Isaacs' book, which says that if you have a $\chi \in Irr(G)$ and a subgroup $K$ then $[\chi_K,\chi_K]=|G:K|$ if and only if $\chi$ vanishes outside $K$.
Now assume $H$ is abelian, $\chi \in Irr(G)$ with $\chi(1)=|G:H|$. By $(*)$ we must have that $\chi$ vanishes off $H$. Look at the set $\{g \in G: \chi(g) \neq 0 \}$. This set is normal (since $\chi(g)=\chi(g^x)$ for any $x \in G$) and is contained in $H$. Hence $N=\langle g \in G: \chi(g) \neq 0 \rangle$ is an abelian normal subgroup of $G$ contained in $H$. Observe that if $h \in H-N$, then we must have $\chi(h)=0$. So $\chi$ vanishes even off $N$, and by the Lemma (2.29) in the other direction and applying (b) to the subgroup $N$, we must have $|G:N|=[\chi_N,\chi_N]\leq \chi(1)=|G:H|$. Hence $|H| \leq |N|$ and we must have $H=N$ and you are done.
Note (February 6th) Returning to my answer above I discovered a flaw in the argument: it is in the last equation, after “we must have” - specifically $[\chi_N,\chi_N] \leq \chi(1)$ is not correct. After some thoughts, I suspected that $(c)$ above is not true at all. As a matter of fact Prof. Derek Holt was very kind to provide me with a counterexample: there is a group of order $32$, SmallGroup(32,6) (notation in Magma or in GAP). It has an abelian subgroup $H \cong C_2 \times C_4$, hence $|G:H|=4$. This subgroup is not normal and $G$ has a $\chi \in Irr(G)$ with $\chi(1)=4$. In this counterexample $|N|=2$.

My fault analysis led me to the following.
Theorem A Let $H \leq G$ be abelian, $\chi \in Irr(G)$ with $|G:H|=\chi(1)$ a prime number, then $H \unlhd G$ and $\chi$ is monomial (induced by a linear character). 
Proof Since $H$ is abelian, $\chi(1) \leq [\chi_H,\chi_H] \leq |G:H|$ and the last inequality is equality if and only if $\chi$ vanishes outside $H$. What about the first inequality? Well, $\chi_H=\sum_{\lambda \in Irr(H)}a_\lambda \lambda$, where the $\lambda$'s are all linear and the integers $a_\lambda \geq 0$. $\chi(1)=\sum_{\lambda \in Irr(H)}a_\lambda$ and $[\chi_H,\chi_H]=\sum_{\lambda \in Irr(H)}a^2_\lambda$. So we see that $\chi(1)=[\chi_H,\chi_H]$ if and only if $a_\lambda \in \{0,1\}$. In that case, $\chi_H=\lambda_1 + \cdots + \lambda _{|G:H|}$. Observe that for these $\lambda$'s by Frobenius Reciprocity, $[\chi_H,\lambda_i]=[\chi,\lambda_i^G]=1$ and it follows that $\chi=\lambda_i^G$ for all $i=1, \cdots ,|G:H|$. Hence, $\chi$ is a monomial character.
Next we concentrate on the normal subgroup $N=\langle g \in G: \chi(g) \neq 0 \rangle$ as defined above, and we apply Clifford Theory to this:
$\chi_N=e\sum_{i=1}^t\mu_i$, where $t=|G:I_G(\mu)|$, the index of the inertia subgroup of the (linear) $\mu_1=\mu$ and $e$ is a positive integer. We already remarked above that $\chi$ vanishes outside $N$, hence $[\chi_N,\chi_N]=e^2t=|G:N|$. And also, $\chi(1)=et=|G:H|$. It follows that $|H:N|=e$.
Now if $|G:H|=p$, $p$ a prime, then $p=et$, hence we have two cases: $e=1$ and $t=p$, or $e=p$ and $t=1$. In the first case we get $|H:N|=1$, that is $H=N$ and hence $H$ is normal. The latter case yields $|G:N|=p^2$, so $G/N$ is a group of order $p^2$ whence abelian, and again $H$ must be normal.$\square$

Note (February 12th) There is a generalization of the above theorem, that I can prove now. It also follows from Udo Riese's results.
Theorem B Let $H \leq G$ be an abelian maximal subgroup, $\chi \in Irr(G)$ with $|G:H|=\chi(1)$, then $H \unlhd G$ and $\chi$ is monomial (induced by a linear character). 
Proof The fact that $\chi$ is monomial runs exactly in the same way as in the proof of Theorem A.
Now I use the 2.11 Theorem from Marty Isaacs' book Finite Group Theory. The proof of this Theorem is based on the famous Zipper Lemma of Helmut Wielandt. The subgroup $H$ satisfies the conditions of this Theorem (caution: in the aforementioned book our $H$ here is called $A$), namely $H$ and $G$ are the only two subgroups between $H$ and $G$ by the maximality of $H$, and we need to check that $|G:H|^2 \leq |G:Z(G)|$. But this trivially true since $|G:H|^2=\chi(1)^2 \leq |G:Z(G)|$ (for the last inequality, this is easy to prove or see (2.30) Lemma in Character Theory of Finite Groups above). By the 2.11 Theorem, it follows $H \subseteq F(G)$, the Fitting subgroup of $G$. Hence either $H=F(G)$, and then $H$ is normal, even characteristic. Or, $G=F(G)$, meaning $G$ is nilpotent and it is well-known that maximal subgroups of nilpotent groups are normal. $\square$

Final remark A famous theorem of I.N. Herstein (1957) asserts that if $G$ has an abelian maximal subgroup, $G$ must be solvable, see here.
