Problem 3.8 from Martin Isaacs' Character theory of Finite groups. Let $\chi$ be a (possible reducible) character of G which is constant on G-{1}. Show that $\chi=a1_{G}+b\rho_{G}$  where a,b $\in \mathbb{Z}$ and $\rho$ is a regular character of G. Also show that if G$\neq ker \chi$, then $\chi(1) \geq |G|-1$. Here $ker\chi$ = {g$\in$G|$\chi(g)=\chi(1)$} which is basically same as $ker \phi$ where $\phi$ is the representation that affords $\chi$.
My approach: Since $\chi$ is constant on G-{1} let $\chi(g)$=a for all g$\neq$1, where a$\in \mathbb{C}$ is some algebraic integer. Since $\rho_{G}$ is the regular character $\rho_{G}(g)$=0 for all g$\neq$1 and $\rho_{G}(1)$=|G|. Let $\chi(1)$=d, that is, the degree of the representation. Let b=(d-a)/|G|, then clearly $\chi=a1_{G}+b\rho_{G}$ for a,b $\in \mathbb{C}$. Now I need to prove that a,b $\in \mathbb{Z}$. Let $\chi_{1}, \chi_{2},...,\chi_{s}$ be all the irreducible characters of G with degrees $d_1,d_2,...,d_s$ respectively. Let also that $\chi_{1}$ is the trivial character that is $1_{G}$. Then since these forms a basis of the set of all class functions and we know how to decompose the regular character we have $\chi$=a$1_{G}$ + b$\sum_{i=1}^{s} d_{i}\chi_{i}$ which gives $\chi=(a+b)1_{G} + b\sum_{i=2}^{s} d_{i}\chi_{i}$. This is the unique linear combination and since $\chi$ is a character each coefficient must be non-negative integer. This gives a+b=a+$\frac{d-a}{|G|}$ is a non-negative integer and also $\frac{d-a}{|G|}.d_{i}$ (for i=2 to s) are non-negative integers. How do I conclude from here that a and b are integer. For the second part also some hint will be appreciated. Thanks in advance!!!
 A: Theorem Let $\phi$ be a possible reducible character (over the complex numbers) of the (non-trivial) group $G$, being constant on $G-\{1\}$. Then the following hold true. 
(a) $\phi=a1_G+b\rho$, where $a$ is an integer and $b$ is a non-negative integer, $1_G$ is the principal character and $\rho$ is the regular character of $G$.
(b) If $ker(\phi)$ is a proper subgroup of $G$, then $\phi(1) \geq |G|-1$.
Proof The proof relies on the fact that the values of inner products of characters are non-negative integers and that character values are algebraic integers, see also I.M. Isaacs CTFG (2.8)Theorem, (2.17)Corollary, (3.2)Lemma and (3.6)Corollary.
Since $\phi$ is a character, it is a linear combination of irreducible characters, that is, $\phi=\sum_{\chi \in Irr(G)}a_{\chi}\chi$, where $a_{\chi} \in \mathbb{Z}_{\geq 0}$. Let us put $\phi(g)=a$ for all $g \in G-\{1\}$. Note that $a \in \mathbb{A}$, the ring of the algebraic integers. We will argue that in fact $a$ is an integer. 
Let us compute the coefficients $a_{\chi}$.
For the principal character $a_{1_G}=[\phi, 1_G]=\frac{1}{|G|}\sum_{g \in G}\phi(g)=\frac{1}{|G|}(\phi(1)+(|G|-1)a)$, which is a non-negative integer, whence $\color{magenta}{\phi(1)+(|G|-1)a}$ is a non-negative integer! Note that $a=\frac{|G|a_{1_G}-\phi(1)}{|G|-1} \in \mathbb{A} \cap \mathbb{Q}=\mathbb{Z}$. So $\color{blue}{a_{1_G}}= \frac{1}{|G|}(\phi(1)+(|G|-1)a)=\frac{\phi(1)-a}{|G|}+a=\color{blue}{a+b}$, with $\color{orange}{b=\frac{\phi(1)-a}{|G|}}$. Since $a$ and $a_{1_G}$ are integers, also $b$ must be an integer.
Now let $\chi \in Irr(G)$ be a non-principal character. This implies that $0=[\chi,1_G]=\frac{1}{|G|}\sum_{g \in G}\chi(g)$. So,$\sum_{g \in G-\{1\}}\chi(g)=-\chi(1).$ But then $\color{red}{a_{\chi}}= [\phi,\chi]=\frac{1}{|G|}\sum_{g \in G}\chi(g)\overline{\phi(g)}=\frac{1}{|G|}\sum_{g \in G}\chi(g)\phi(g)=\frac{1}{|G|}(\chi(1)\phi(1)+\sum_{g \in G-\{1\}}\chi(g)a)=\frac{1}{|G|}(\chi(1)\phi(1)-a\chi(1))=\color{red}{\chi(1)b}.$ Note that from this it follows that $b$ is a non-negative integer!
We conclude that
$$\phi=\sum_{\chi \in Irr(G)}a_{\chi}\chi=a_{1_G}1_G+\sum_{\chi \in Irr(G)-{1_G}}a_{\chi}\chi=(a+b)1_G+\sum_{\chi \in Irr(G)-{1_G}}\chi(1)b\chi=$$$$=a1_G+b\sum_{\chi \in Irr(G)}\chi(1)\chi=a1_G+b\rho.$$ This proves (a).
Now to prove (b) we observe that from the last formula it follows that $$b \neq0 \iff \phi(1) \neq a \iff G \gt ker(\phi)$$ and $$\phi(1)=a+b|G|.$$ Now assume that $ker(\phi) \lt G$, which is equivalent to $b \geq 1$. Since $a$ is an integer, we have two cases:


*

*$a \leq -1$ or

*$a \geq 0$.
In the first case, we are using the fact that in the beginning we observed that $\phi(1)+(|G|-1)a $ is a non-negative integer. Hence, $\phi(1)+(|G|-1)a \geq 0 \iff \phi(1) \geq(|G|-1)(-a) \geq |G|-1$. In the second case, we are using $\phi(1)=a+b|G| \geq b|G| \geq |G| \gt |G|-1 \text{   }\square$.
