Character as sum with regular representation Suppose $G$ is a group and $\chi$ is a character of $G$ with $\chi(g_1)=\chi(g_2)$ for all non-identity $g_1,g_2 \in G$, and let $\chi_{reg}$ denote the regular representation character. I read that $\chi=\alpha1_G+\beta\chi_{reg}$ for some complex $\alpha, \beta$ and that $\alpha+\beta, \alpha+\beta|G|, \alpha, \beta$ are all integers, but I am confused as to why this is true. 
I think that $\chi(g_1)=\chi(g_2)$ for all non-identity $g_1,g_2 \in G$ means that all non-id elements are in the same conjugacy class so that there are only two conjugacy classes in $G$, but I am lost from there. Any help would be appreciated. 
 A: So this question is really old, but I have a solution if you're interested.
Showing that $\alpha + \beta|G|, \alpha + \beta \in \mathbb{Z}$ is pretty straightforward, since $\chi(e)= \alpha + \beta |G|$ and $\langle \chi_{tr}, \chi \rangle= \alpha + \beta$. Once you have this, you can show that $\beta \in \mathbb{Z}$, and then it follows that $\alpha \in \mathbb{Z}$. First notice that $\beta(|G|-1) \in \mathbb{Z}$. So, since $|G|-1 \in \mathbb{Z}$, we know that $\beta \in \mathbb{Q}$. Write $\beta = \frac pq$ with $(p, q) = 1$. If we show that $\beta$ is an algebraic integer, it will follow that $\beta \in \mathbb{Z}$.
Consider the irreducible representations of $G$, with corresponding characters $\chi_1, \dots, \chi_r$ where $\chi_1 = \chi_{tr}$. For $k \neq 1$, note that
$$
\langle \chi, \chi_k \rangle = \alpha \langle \chi_{tr}, \chi_k \rangle + \beta \langle \chi_{reg}, \chi_k \rangle = \beta \chi_k(e),
$$
since $\chi_k$ is irreducible. Since $\langle \chi, \chi_k \rangle \in \mathbb{Z}$, we have $\beta \chi_k(e) \in \mathbb{Z}$. It follows that $q \mid \chi_k(e)$, so $\chi_k(e) = m_k q$ for some $m_k \in \mathbb{Z}$.
For $g \neq e$ we have
$
\sum_{k=1}^r \chi_k(e)\chi_k(g) = 0
$, 
so
$$
- \beta = \beta \sum_{k=2}^r \chi_k(e) \chi_k(g)= \beta\sum_{k=2}^r m_k q \cdot \chi_k(g) = p \sum_{k=2}^r m_k \chi_k(g).
$$
For $g \neq e$ we know that $\chi_k(g)$ is an algebraic integer. Thus, since the algebraic integers form a ring, $\beta$ is an algebraic integer. 
