Irreducible character of degree greater than one takes value zero on some conjugacy class

Statement

Suppose $G$ is a finite group and $\chi$ is the character of an irreducible linear representation of $G$ over $\mathbb {C}$, such that the degree of the representation (and hence, of $\chi$) is greater than one. Then, there exists an element $g\in G$ (and hence, a Conjugacy class (?)) such that $\chi (g)=0$.

Related facts

• Zero-or-scalar lemma
• Conjugacy class of more than average size has character value zero for some irreducible character

Examples

Representation	Degree	Conjugacy class(es) where the character is zero	Group	Linear representation theory information
standard representation of symmetric group:S3	2	$(1,2)$ -- class of 2-transpositions	symmetric group:S3	linear representation theory of symmetric group:S3
faithful irreducible representation of dihedral group:D8	2	all the conjugacy classes of elements outside the center.	dihedral group:D8	linear representation theory of dihedral group:D8
faithful irreducible representation of quaternion group	2	all the conjugacy classes of elements outside the center.	quaternion group	linear representation theory of quaternion group
standard representation of symmetric group:S4	3	$(1,2,3)$ -- class of 3-cycles	symmetric group:S4	linear representation theory of symmetric group:S4

Facts used

1. Character orthogonality theorem
2. Sufficiently large implies splitting: In particular, this exhibits a splitting field of characteristic zero that is a finite cyclotomic extension of the rationals.
3. Cauchy-Schwartz inequality
4. Characters are algebraic integers

Proof

Given: A finite group $G$, $\chi$ the character of an irreducible representation of degree greater than $1$ of the group $G$. $e$ is the identity element, $d=\chi (e)$ is the degree, so $d>1$.

To prove: There exists $g\in G\setminus \{e\}$ such that $\chi (g)=0$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$\sum _{g\in G}|\chi (g)|^{2}=|G|$	Fact (1)			Use Fact (1), and move the $1/|G|$ to the other side.
2	$d^{2}+\sum _{g\in G\setminus \{e\}}|\chi (g)|^{2}=|G|$			Step (1)	Pull out the term for the identity element.
3	$\sum _{g\in G\setminus \{e\}}|\chi (g)|^{2}=|G|-d^{2}<|G|-1$		$d>1$, i.e., the representation has degree more than one	Step (2)	Rearrange Step (2), use $d>1$.
4	Suppose $K$ is a finite degree cyclotomic extension of $\mathbb {Q}$ that is a splitting field for $G$. Let $H$ be the Galois group of the field extension $K/\mathbb {Q}$. Then, $H$ acts on the set of irreducible representations, with an automorphism $\sigma$ acting by: $\varphi \mapsto (g\mapsto \sigma (\varphi (g))$	Fact (2)			Fact (2) guarantees the existence of such a $K$.
5	$\sum _{g\in G\setminus \{e\}}\left[\prod _{\sigma \in H}|\sigma (\chi (g))|\right]^{2}\leq \prod _{\sigma \in H}\left[\sum _{g\in G\setminus \{e\}}|\sigma (\chi (g))|^{2}\right]^{1/|H|}$	Fact (3)		Steps (3), (4)
6	$\sum _{g\in G\setminus \{e\}}|N_{K}(\chi (g))|^{2}\leq \prod _{\sigma \in H}\left[\sum _{g\in G\setminus \{e\}}|\sigma (\chi (g))|^{2}\right]^{1/|H|}$			Step (5)	[SHOW MORE] The products on the left side of Step (5) are equal to the algebraic norms of the elements $\chi (g)$ in $K$.
7	$\sum _{g\in G\setminus \{e\}}|\sigma (\chi (g))|^{2}=|G|-d^{2}<|G|-1$ for every $\sigma \in H$			Steps (2), (4)	[SHOW MORE] We apply the reasoning of Step (2) to Failed to parse (syntax error): {\displaystyle \\sigma \circ \chi} in place of $\chi$, knowing that if $\chi$ is the character of $\varphi$, then $\sigma \circ \chi$ is the character of $\sigma \circ \varphi$, and thus the same reasoning applies.
8	$\prod _{\sigma \in H}\left[\sum _{g\in G\setminus \{e\}}|\sigma (\chi (g))|^{2}\right]^{1/|H|}<|G|-1$			Step (7)	[SHOW MORE] Traverse Step (7) for all $\sigma \in H$, multiply all and raise both sides to the power of $1/|H|$.
9	$\sum _{g\in G\setminus \{e\}}|N_{K}(\chi (g))|^{2}<|G|-1$			Steps (6), (8)	Step-combination direct
10	Each $N_{K}(\chi (g))$ is an integer, so each $|N_{K}(\chi (g))|^{2}$ is a nonnegative integer.	Fact (4)		Step (4) (definition of $K$)	[SHOW MORE] Since $\chi (g)$ is an algebraic integer, a product of its conjugates is also an algebraic integer, and since that's also rational, it must be an integer.
11	For some $g\in G\setminus \{e\}$, $N_{K}(\chi (g))=0$			Steps (9), (10)	[SHOW MORE] Each of the terms $|N_{K}(\chi (g))|^{2}$ is a nonnegative integer, but if all of them are positive, the left expression becomes at least $|G|-1$, leading to a contradiction with Step (9). Thus, , so $\chi (g)$ is zero. This completes the proof.
12	For some $g\in G\setminus \{e\}$, $\chi (g)=0$			Step (11)	Follows from Step (11) and the observation that $N_{K}(x)=0\iff x=0$.