Let $\chi$ be a character of a finite group $G$ which is constant on $G\backslash\{1\}$. Show that $\chi = a1_G + b\rho_G$ where $a,b\in\mathbb{Z}$.

Question

I have been working on this equality for a while, and am a bit stuck.

In the case that $\chi$ is identically $0$ on $G\backslash\{1\}$, I have shown that $\chi$ must be an integer multiple of $\rho_{G}$, the regular character.

I had hoped that this would aid me, though it has not seemed to help.

I have also taken an approach of computing $\left<\chi,1_{G}\right>$, showing it is non zero and thus $1_{G}$ appears an integer number of times in $\chi$. I then attempted to show that $b$ is an integer, based on what I had computed for $\left<\chi,1_{G}\right>$.

Any help would be appreciated.

Answer 1

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

Answer 2

The trivial character $\chi_0$ is $1$ for all arguments. The regular rep has character

$$ \chi_G(g)=\begin{cases} |G| & g=e \\ 0 & g\ne e \end{cases} $$

The regular representation is an example of a permutation representation. For any permutation representation, the group elements correspond to permutation matrices, so the character of a group element corresponds to the trace of the corresponding permutation matrix, which is the number of $1$s along its diagonal, which equals the number of fixed points of the permutation. In the regular representation, a group element turns into a permutation of the group via multiplication (a la Cayley's theorem); multiplying-by-$e$ has all $|G|$ group elements as fixed points, whereas multiplying-by-$g$ has no elements as fixed points.

Therefore the linear combination $a\chi_0+b\chi_G$ of the trivial and regular characters is

$$ a\chi_0(g)+b\chi_G(g)=\begin{cases} a+|G|b & g=e \\ a & g\ne e \end{cases} $$

Now, if we have some character $\chi$ of the form

$$ \chi(g)=\begin{cases} u & g=e \\ v & g\ne e \end{cases} $$

Then we can solve $a=v$ and $b=\frac{u-v}{|G|}$ so that $\chi=v\chi_0+\frac{u-v}{|G|}\chi_G$.

Suppose the irreps of $G$ are $\chi_0,\chi_1,\cdots,\chi_{k-1}$ with degrees $d_0,d_1,\cdots,d_{k-1}$. For each $i\ge0$ the irrep $\chi_i$ has multiplicity $d_i$ within $\chi_G$. Within $\chi$ we find $\chi_0$ has multiplicity $v+\frac{u-v}{|G|}$ and $\chi_i$ for $i>0$ has multiplicity $\frac{u-v}{|G|}d_i$. This means any $\mathbb{Z}$-linear combination of the $\frac{u-v}{|G|}d_i$s is an integer - by Bezout's theorem this implies $\frac{u-v}{|G|}d$ is an integer, where $d=\gcd(d_1,\cdots,d_{k-1})$. All the $d_i$s are factors of $|G|$, so $d$ is too, but $d_1^2+\cdots+d_{k-1}^2=|G|-1$ implies $d$ is also a divisor of $|G|-1$ which is coprime to $|G|$, forcing $d=1$. Therefore, $b=\frac{u-v}{|G|}$ is an integer, forcing $a=v$ to be an integer too.

Answer 3

Here is a proof very similar to the others, but I think slightly simplified. It uses that inner products of characters are integers, and that rational character values are integers. I don't see any easy way to avoid this at the moment.

Let $\chi$ be a character such that $\chi$ is constant on $G\setminus \{1\}$, let $x\in G$ be non-trivial, and write $\alpha=\chi(1)$, $\beta=\chi(x)$. First, we note that $$ \langle \chi,1_G\rangle=\frac{1}{|G|}(\beta(|G|-1)+\alpha)=\beta+\frac{(\alpha-\beta)}{|G|}.$$

Suppose first that this is zero, i.e., $\langle \chi,1_G\rangle=0$. Then $\alpha=-\beta(|G|-1)$. Since $\alpha=\chi(1)$ is an integer, $\beta$ is a rational number, hence an integer, as the only rational character values are integers. Thus $$ \chi-\beta\cdot 1_G$$ is $0$ on $G\setminus \{1\}$, and an integer value on $1$. This is a scalar multiple of the regular character, hence an integer multiple $\alpha/|G|\cdot \rho_G$ of the regular character, as you said you have proved. Thus $\chi=\alpha/|G|\cdot \rho_G+\beta\cdot 1_G$, as needed.

Finally, for general $\chi$, $\chi=\langle \chi,1_G\rangle+\chi_0$, where $\chi_0$ satisfies the hypotheses of the question and has inner product zero. Thus $\chi_0$ is an integer linear combination of $1_G$ and $\rho_G$, whence so is $\chi$.