# Why $χ$ is linear combination of identity representation and regular representation with integer coefficients

abstract-algebracharactersgroup-theoryrepresentation-theory

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.

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

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