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Let $G$ be a finite group. The regular $\chi_{reg}$ of $G$ is the character corresponding to representation of $G$ on $\mathbb{C}[G]$ with action by (left) multiplication to basis elements of $G$.

It is well known that $\chi_{reg}$ is zero at all $g\in G$ except $g=1$. Also the degree of this character (i.e. value at $1$) is $|G|$.

There is an exercise in Isaacs' character theory about similar kind of character:

Fact: Let $G$ be a finite group and $\chi$ a character of $G$ such that $\chi$ vanishes on $G-\{1\}$. Then $|G|$ divides degree of character.

Question: In the above fact, can we also say that $\chi$ contains $\chi_{reg}$ as a component? (i.e., is $\chi$ equal to sum of $\chi_{reg}$ and some other character of $G$?


Note: In above discussion, character of $G$ means character of a complex (possibly reducible) representation of $G$, and not virtual character, or class function etc.

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    $\begingroup$ Won't these follow from the basic properties of the inner product? The Fact is true because the inner product of $\chi$ and the trivial character must be a natural number $m$. Therefore $\chi$ is a sum of $m$ copies of $\chi_{reg}$, no? $\endgroup$ – Jyrki Lahtonen Apr 30 '17 at 5:00
  • $\begingroup$ As Lord Shark also explained, we get that $\chi=m\chi_{reg}$, so $\chi$ is a sum of $m$ copies of $\chi_{reg}$. $\endgroup$ – Jyrki Lahtonen Apr 30 '17 at 5:35
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    $\begingroup$ @JyrkiLahtonen Although this is not difficult to prove, I am not really following either your explanation or Lord Shark's. $\endgroup$ – Derek Holt Apr 30 '17 at 10:07
  • $\begingroup$ @DerekHolt What have I missed? If $\chi_1$ is the trivial character, then we get that $$\langle\chi,\chi_1\rangle=\frac{\chi(1)}{|G|}$$ must be a natural number $m$, and hence $\chi=m\chi_{reg}$? Or, are you saying that this exercise comes before we know the values of $\chi_{reg}$, and the linear indepence of characters (plus whatever else I used tacitly). $\endgroup$ – Jyrki Lahtonen Apr 30 '17 at 12:27
  • $\begingroup$ Beginner, $\chi_{reg}$ is not irreducible, so it is a bit unclear to discuss how many times it appears in another representation, because the irreducible components of $\chi_{reg}$ might appear in $\chi$ with different multiplicities. All I'm saying is that because $\chi=m\chi_{reg}$, the representation affording $\chi$ is isomorphic to $\Bbb{C}[G]^m$. $\endgroup$ – Jyrki Lahtonen Apr 30 '17 at 13:12
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This character $\chi$ is a multiple of the character of the regular representation: $\chi=a\chi_{reg}$. A priori $a\in\Bbb Q$, but the regular representation contains one copy of the trivial representation. So $\chi$ contains $a$ copies of the trivial representation, so $a\in\Bbb Z$. Therefore the degree of $\chi$ is $a|G|$.

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  • $\begingroup$ I know, I was missing the obvious! I'll delete my previous comment. $\endgroup$ – Derek Holt Apr 30 '17 at 16:22

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