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Let $G$ be a finite group and $S\subseteq G$. Let $\rho$ be an irreducible representation of $G$ and $\chi$ be its (irreducible) character. Define $$\rho(S):=\sum_{s\in S}\rho(s),$$ $$\chi(S):=\sum_{s\in S}\chi(s).$$ By Schur's lemma, it can be easily checked that, if $S$ is normal in $G$, $\rho(S)=\lambda I$, for some $\lambda\in\mathbb{C}$. Moreover, it follows that $\lambda=\frac{\chi(S)}{deg(\chi)}$.

I think the converse is also true, i.e., if $\rho(S)$ is a scalar matrix, for some irreducible, non-trivial representation $\rho$ of $G$, then $S$ is a normal subset of $G$. I was thinking of showing that the characteristic function $\alpha:G\to\mathbb{C}$ defined as, $$\alpha(x)=\begin{cases} 1, \text{if $x\in S$}\\ 0, \text{if $x\notin S$} \end{cases}$$ is a class function on $G$. Then it will essentially follow that $S$ is a normal subset of $G$. Is this approach correct? Or is my surmise wrong?

Lucenaposition
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SPDR
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    If $\rho$ is the trivial representation, then $\rho(S)=|S|\cdot I$ for any set $S$. – Hjalmar Rosengren Sep 02 '25 at 13:09
  • You need some more conditions. For example, as you have it now, every subset of such a set $S$ meets your requirements... – Steve D Sep 02 '25 at 13:15
  • @HjalmarRosengren Yes, I should have mentioned $\rho$ to be non-trivial. Just edited my question. – SPDR Sep 02 '25 at 13:31
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    @SteveD Sorry, I didn't get your point. Did you mean that if $\rho(S)=\lambda I$, then $\rho(T)=\mu I$, for every $T\subseteq S$? – SPDR Sep 02 '25 at 13:37

1 Answers1

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Asking for $\rho$ to be nontrivial is not a strong enough hypothesis, because it can still have a kernel, and then any subset of the kernel has this property. For example we can consider the sign representation $\rho : S_n \to \{ \pm 1 \}$ of the symmetric groups, whose kernel is $A_n$, then take any subset of $A_n$ which is not a union of conjugacy classes.

More interestingly, asking for $\rho$ to also be faithful is still not enough (because it won't be a faithful representation of the group algebra $\mathbb{C}[G]$). For example, the dihedral group $D_n$ has a faithful irreducible ($n \ge 3$) $2$-dimensional representation which, when $n$ is even, contains the matrix $\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$. So if $g \in D_n$ then $\rho(g) + (- \rho(g)) = 0$ is a sum of (the image of) two elements of $D_n$, but $\{ g, -g \}$ is usually not the conjugacy class of $g$, e.g. if $g$ is a counterclockwise rotation by $\frac{2 \pi k}{n}$ then $-g$ is a rotation by $\frac{2 \pi k}{n} + \pi$, but $g$ is conjugate to $g^{-1}$ which is a rotation by $- \frac{2 \pi k}{n}$.

What is true is the following:

If $S$ is a subset of $G$ such that, for every irreducible representation $\rho$, $\rho(S)$ is a scalar multiple of the identity, then $S$ is a union of conjugacy classes.

More generally, if $z \in \mathbb{C}[G]$ is an element of the group algebra such that, for every irreducible representation $\rho$, $\rho(z)$ is a scalar multiple of the identity, then $z$ is a sum over conjugacy classes. Equivalently, writing $z = \sum z_g g$, so that $\rho(z) = \sum z_g \rho(g)$, the function $g \mapsto z_g$ is a class function.

(Sketch: once we know that irreducible representations are jointly faithful / separate points in the sense that if $\rho(z) = 0$ for all irreducible $\rho$ then $z = 0$, this condition implies that $z$ lies in the center $Z(\mathbb{C}[G])$ of the group algebra. Then we compute that a central element of $\mathbb{C}[G]$ is exactly a fixed point under the action of conjugation by elements of $G$, which is exactly a sum over conjugacy classes.)

Qiaochu Yuan
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  • Could you please explain the "jointly faithful/ separate points" thing in a bit detail? I mean how does it follow that $z=0$? – SPDR Sep 04 '25 at 13:48
  • @SPDR: I'm not sure what you're asking. Are you asking what this condition means, or how to prove that this condition holds? If the latter, this follows from Maschke's theorem (https://en.wikipedia.org/wiki/Maschke%27s_theorem) together with the Artin-Wedderburn theorem (https://en.wikipedia.org/wiki/Wedderburn%E2%80%93Artin_theorem). – Qiaochu Yuan Sep 04 '25 at 16:55