Zero-or-scalar lemma

Statement

Let $G$ be a finite group and $\varphi$ a nontrivial Irreducible linear representation (?) of $G$ over $\mathbb {C}$. Let $g\in G$, such that the size of the conjugacy class of $G$ is relatively prime to the degree of $\varphi$. Then, either $\varphi (g)$ is a scalar or $\chi (g)=0$.

Facts used

1. Size-degree-weighted characters are algebraic integers
2. Characters are algebraic integers
3. Element of finite order is semisimple and eigenvalues are roots of unity

Proof

Given: A finite group $G$, a nontrivial irreducible linear representation $\varphi$ of $G$ over $\mathbb {C}$ with character $\chi$. An element $g\in G$ with conjugacy class $C$. The degree of $\varphi$ and the size of $C$ are relatively prime.

To prove: Either $\chi (g)=0$ or $\varphi (g)$ is a scalar.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The number ${\frac {|C|\chi (g)}{\chi (1)}}$ is an algebraic integer.	Fact (1)	$G$ is finite, $\varphi$ is an irreducible representation of $G$ over $\mathbb {C}$ with character $\chi$		Given+Fact direct
2	There exist integers $a$ and $b$ such that $\!a|C|+b\chi (1)=1$		$|C|$ and $\chi (1)$ (the degree of $\varphi$) are relatively prime.		By definition of relatively prime.
3	We get ${\frac {a|C|\chi (g)}{\chi (1)}}+b\chi (g)={\frac {\chi (g)}{\chi (1)}}$			Step (2)	Multiply both sides of Step (2) by $\chi (g)/\chi (1)$.
4	The expression ${\frac {a|C|\chi (g)}{\chi (1)}}+b\chi (g)$ gives an algebraic integer.	Fact (2)		Step (1)	[SHOW MORE] By Step (1), $|C|\chi (g)/\chi (1)$ is an algebraic integer. By fact (2), $\chi (g)$ is also an algebraic integer. Since $a,b$ are (rational) integers, the sum is thus also an algebraic integer.
5	$\chi (g)/\chi (1)$ is an algebraic integer.			Steps (3), (4)	[SHOW MORE] By Step (4), the left side of the expression of Step (3) is an algebraic integer. Hence, so is the right side.
6	$\chi (g)$ is the sum of $\chi (1)$ many roots of unity (not necessarily all distinct), namely, the eigenvalues of the corresponding element $\varphi (g)$..	Fact (3)	$G$ is finite.		[SHOW MORE] Since $G$ is finite, $g$ has finite order, hence so does $\varphi (g)$. Thus, by fact (3), it is semisimple and all its eigenvalues ($\chi (1)$ many of them, since that is the dimension of the vector space on which it is acting) are roots of unity. $\chi (g)$ is the sum of these.
7	Every algebraic conjugate of $\chi (g)$ is also a sum of $\chi (1)$ roots of unity.			Step (6)	[SHOW MORE] Any Galois automorphism on $\chi (g)$ must send each of the roots of unity that add up to it to other roots of unity, which now add up to its image under the automorphism.
8	Every algebraic conjugate of $\chi (g)/\chi (1)$ has modulus less than or equal to $1$.			Step (7)	[SHOW MORE] By Step (7) and the triangle inequality, the modulus of any algebraic conjugate of $\chi (g)$ is less than or equal to $\chi (1)$. Dividing by $\chi (1)$, we get the desired conclusion.
9	The modulus of the algebraic norm of $\chi (g)/\chi (1)$ in a Galois extension containing it is either 0 or 1.			Steps (5), (8)	[SHOW MORE] By definition, the algebraic norm is the product, under all automorphisms of the Galois extension, of the images under those automorphisms. Taking modulus and using Step (8), we see that that its modulus is the product of numbers between 0 and 1, hence is between 0 and 1. By Step (5), $\chi (g)/\chi (1)$ is an algebraic integer, so its norm is a rational integer, hence the modulus of the norm is a nonnegative integer. The only possibilities are thus 0 and 1.
10	If the modulus of the algebraic norm of $\chi (g)/\chi (1)$ is $0$, then $\chi (g)=0$				[SHOW MORE] If the algebraic norm has modulus zero, then the algebraic norm is zero, which can happen only if one of the algebraic conjugates of $\chi (g)/\chi (1)$ is zero. But by the definition of algebraic conjugates, any conjugate being zero forces all of them to be zero, so $\chi (g)/\chi (1)$ is zero, so $\chi (g)$ is zero.
11	If the modulus of the algebraic norm of $\chi (g)/\chi (1)$ is $1$, then $\varphi (g)$ is a scalar matrix.			Steps (6), (8)	[SHOW MORE] By Step (8), every algebraic conjugate has modulus less than or equal to 1. If the product of all of these has modulus 1, then every one of them must have modulus 1. In particular, $\chi (g)/\chi (1)$ has modulus 1, so $|\chi (g)|=\chi (1)$. By Step (6), $\chi (g)$ is the sum of $\chi (1)$ roots of unity obtained as eigenvalues of $\varphi (g)$. By the triangle inequality, we see that $|\chi (g)|\leq \chi (1)$ and equality holds if and only if all the roots of unity are equal. Thus, all eigenvalues of $\varphi (g)$ are equal, forcing $\varphi (g)$ to be scalar.
12	Either $\chi (g)=0$ or $\varphi (g)$ is scalar.			Steps (9), (10), (11)	Step-combination direct.