Equality of subsets of abelian groups Let $G$ be a finite abelian group, $X$ and $Y$ be two non-empty subsets of $G$ of equal size. Suppose that for each irreducible character $\chi$ of $G$ we have $\sum_{x\in X}\chi(x)=\sum_{y\in Y}\chi(y)$. Is it true that $X=Y$ in general?
 A: $\DeclareMathOperator\Irr{Irr}$You can see this from the fact that for abelian groups, irreducible characters form a $\mathbb{C}$-basis of the space of functions from $G$ to $\mathbb{C}$. These functions correspond bijectively to linear transformations from the group algebra $\mathbb{C}G$ to $\mathbb{C}$, and $\Irr(G)$ is also a basis for the space of such linear transformations. Therefore, $\chi(\sum_{x\in X}x)=\sum_{x\in X}\chi(x)=\sum_{y\in Y}\chi(y)=\chi(\sum_{y\in Y}y)$ for each $\chi\in \Irr(G)$ implies that $f(\sum_{x\in X}x)=f(\sum_{y\in X}y)$ for every linear transformation $f:\mathbb{C}G\rightarrow \mathbb{C}$. This happens if and only if $\sum_{x\in X}x=\sum_{y\in Y}y$.
There is another way to see this. In general, for split finite dimensional algebra $A$ over a field $k$, the common zero set of irreducible characters $\Irr_k(A)$ is $J(A)+[A,A]$, where $J(A)$ is the Jacobson radical of $A$ and $[A,A]$ is the commutator subalgebra of $A$ generated by elements of the form $[a,b]=ab-ba$.
If $G$ is a finite abelian group, then the group algebra $\mathbb{C}G$ is finite dimensional, split, semisimple and commutative. Hence, both $J(A)$ and $[A,A]$ are trivial, and the above theorem says that the common zero set of irreducible complex characters is trivial. In particular, if two elements $a$, $b$ of the group algebra satisfy $\chi(a)=\chi(b)$ for each $\chi\in \Irr(G)$, then $a-b$ is in the common zero set, which is $\{0\}$.
A: More generally, for any finite group $G$, Abelian or not, it is true that $Z(\mathbb{C}G)$, the center of the complex group algebra $\mathbb{C}G$, has a $\mathbb{C}$-basis of mutually orthogonal idempotents, $\{e_{\chi} : \chi \in {\rm Irr}(G) \},$ indexed by the complex irreducible characters of $G$. In other words,we have $e_{\chi}^{2} = e_{\chi}$ for each $\chi$, and $e_{\chi}e_{\mu} = 0$ when $\chi \neq \mu.$  This property ensures that the idempotents $\{ e_{\chi} \}$ are linearly independent
Given an element $X \in Z(\mathbb{C}G),$ we may write $X$ uniquely in the form
$X = \sum_{\chi} a_{\chi}(X) e_{\chi}$ where the $a_{\chi}(X)$ are complex numbers, and we may check that for each $\chi$, the map $X \to a_{\chi}(X)$ is an algebra homomorphism from $Z(\mathbb{C}G)$ to $\mathbb{C}$, using the orthogonality of the idempotents $e_{\chi}$.
Notice that $a_{\chi}(e_{\mu}) = \delta_{\chi, \mu}$ by definition, and that $a_{\chi}(1_{G}) = 1$ for each $\chi.$ By Schur's Lemma, we see that $e_{\chi}$ is represented by a scalar matrix in any complex representation of $G$ affording irreducible character $\chi$, and this matrix must be idempotent, so that $\chi(e_{\chi}) = \chi(1)$
Thus we have $a_{\chi}(e_{\mu}) = \frac{\chi(e_{\mu})}{\chi(1)}$ for all irreducible characters $\chi, \mu.$ Hence we have $a_{\chi}(X) = \frac{\chi(X)}{\chi(1)}$ for all $X \in Z(\mathbb{C}G)$, since $\{e_{\mu} \}$ is a $\mathbb{C}$-basis for $Z(\mathbb{C}G).$
Consequently, we may conclude that two elements $X,Y \in Z(\mathbb{C}G)$ are equal if and only if $\chi(X) = \chi(Y)$ for each complex irreducible character $\chi$ of $G$.
In case $G$ is Abelian, the group algebra $\mathbb{C}G$ is commutative, and is equal to its center $Z(\mathbb{C}G)$. Furthermore, $G$ has $|G|$ complex irreducible characters all of degree $1$, so we obtain that $X,Y \in \mathbb{C}G$ are equal if and only if $\lambda(X) = \lambda(Y)$ for each irreducible (linear) complex character $\lambda$ of $G$.
