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!!!