Let $G$ be a finite group.

And let $χ$ be character of representation of $G$ over $ \Bbb C$, such that there exists some complex number $a$ for all non identity element of $G$, $χ(g)＝a・・・①$.

Then, why $χ$ is linear combination of identity representation and regular representation with **integer** coefficients?

My thought and embarrassment:

From property of character of regular representation, I need to prove

$χ(1)＝c＋d｜G｜(c,d∈ \Bbb Z)$ when $g＝1$,

$χ(g)＝c$ when $g≠1$ for some integer $c,d$.

But from ①, when $g≠1$, $χ(g)＝a$, and $a$ need not to be equal to integer, I'm wondering why $χ(g)$ becomes integer.

## Best Answer

If you try to solve Isaacs' Problem $(3.8)$ ...

TheoremLet $\phi$ 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

principalcharacter and $\rho$ is theregularcharacter of $G$.(b) If $ker(\phi)$ is a proper subgroup of $G$, then $\phi(1) \geq |G|-1$.

ProofThe 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-principalcharacter. 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-negativeinteger!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:

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 $\phi(1)=a+b|G| \geq b|G| \geq |G| \gt |G|-1 \text{ }\square$.