# Prove that $\forall g \in G$ $\exists$ an irreducible non-trivial character $\chi$ of the group $G$ such that $\chi(g)\neq 0$

Let $G$ be a non-trivial finite group. Prove that $\forall g \in G$ $\exists$ an irreducible non-trivial character $\chi$ of the group $G$ such that $\chi(g)\neq 0$

This is my attempt so far.

Let us suppose for a contradiction that $\exists g \in G$ such that for all non-trivial $\chi\in Irr(G)$ we have $\chi(g)=0.$

At this point I am now stuck. I assume that this has something to do with the orthogonality relations but I am unsure how to proceed.

Am I on the right path?

Thank you.

• This character is a multiplicative functional right? – voldemort Jan 30 '14 at 21:51
• Isn't a character a homomorphism $\chi\colon G\to\mathbb C^\times$? And isn't $0\notin\mathbb C^\times$? – Hagen von Eitzen Jan 30 '14 at 22:00
• @HagenvonEitzen: See here for a definition: character. – Dejan Govc Jan 30 '14 at 22:13
• @Hagen: for non-abelian groups, homomorphisms to $\mathbb C^{\times}$ are not enough. Indeed, they kill all commutators. – GEdgar Jan 30 '14 at 22:48

Yes, you are on the right track. Simply use the column orthogonality relations: the dot product of your column (which contains only zeros and a single $1$) with any other column in the character table (for example the one belonging to $1$) should be zero, which leads immeadiately to a contradiction.
• Another way of saying it, $\sum \chi(g) = 1$ means $g=1$ and $|G|=1$. – Jack Schmidt Jan 30 '14 at 22:26
Another way is to look at the regular representation. $g \ne e$ has trace zero on this, but there is a copy of the trivial representation given by the sum of all the group elements which we can quotient by to get a representation where $g$ has character $-1$.