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> [!proposition] Proposition. ([[character table test for simplicity]])
> Let $G$ be a finite [[group]] with [[irreducible group representation]]s $\{ \rho_{j} \}_{j=1}^{\# \text{CCs}}$. Then $G$ is *not* [[simple group|simple]] if and only if its [[character table]] has a cell in row $\ell$ not corresponding to the [[trivial group representation]] and column not corresponding to $[e]$ which takes on the value $\dim \rho$.
> \
> In other words, $G$ is simple if and only if all nontrivial irreducible kernels have trivial kernel.
> [!basicexample]
> | $C_{3}$ | $[e]$ | $[x]$ | $[x^2]$ |
|-----|-----|-------|-------|
| $\chi_{\mathbb{1}}$ | 1 | 1 | 1 |
| $\chi$ | 1 | $\omega$ | $\omega^{2}$ |
| $\chi^{2}$ | 1 | $\omega^{2}$ | $\omega$ |
$C_{3}$ is simple because no nontrivial cell takes on the value $1$.
>
| $C_{4}$ | $[e]$ | $[x]$ | $[x^2]$ | $[x^3]$ |
|-----|-----|-------|-------|-------|
| $\chi_{\mathbb{1}}$ | 1 | 1 | 1 | 1 |
| $\chi$ | 1 | $-i$ | $-1$ | $i$ |
| $\chi^{2}$ | 1 | $-1$ | $1$ | $-1$ |
| $\chi^{3}$ | 1 | $i$ | $-1$ | $i$ |
$C_{4}$ is not simple because $[x^{2}]$ takes on the value $1$ in the row corresponding to $\chi^{2}$ and $\dim \chi^{2} = 1$.
> [!proof]- Proof. ([[character table test for simplicity]])
> $\to$. Suppose there exists a nontrivial [[irreducible group representation|irrep]] $\rho$ whose row in the [[character table]] has a cell in some column $k$ not corresponding to $[e]$ that takes on the value $\dim \rho$. Then there exists $g \in G$ with $g \neq e$ such that $\text{tr } \rho_{g}= \dim \rho$. [[trace of a linear operator|That is]], the [[eigenvalue|eigenvalues]] of $\rho_{g}$ sum to $\dim \rho$; since each has unit modulus ([[eigenvalues of order-m matrix in GL_n(C) are mth roots of unity|as roots of unity]]) we must in fact have $\text{eig}(\rho_{g}) \equiv 1$. Now, since $\rho_{g}$ is [[diagonalizable]] (all [[matrices of finite order in GL_n(C) are diagonalizable]]), we must have $\rho_{g}=I$ by [[diagonalizable matrix has eigenvalues all 1 iff is identity]]. But this means $g \in \ker \rho$ and thus $\ker \rho$ is nontrivial. Also $\ker \rho \neq G$ because $\rho \neq \rho_{\mathbb{1}}$. Hence $\ker \rho$ [[kernel iff normal subgroup|is a]] proper nontrival [[normal subgroup]] of $G$ and therefore $G$ is not simple.
>
> $\leftarrow.$ Conversely suppose $G$ is not [[simple group|simple]]: fix a proper nontrival [[normal subgroup]] $N \trianglelefteq G$. $N$ is the [[kernel]] of the [[kernel iff normal subgroup|natural projection homomorphism]] $\pi: G \to G / N$. Let $\rho^{(\pi)}$ be a [[group representation|representation]] of $G / N$. Then define $\rho': G \to \text{GL}_{}(V)$ to be given by $\rho'(g):=\rho^{(\pi)}(gN)$. For all $g \in N$, $\rho'(g)=I$ and thus $g \in \ker \rho'$. Using [[Maschke's Theorem]] we can break $\rho'$ into a [[direct sum of representations|direct sum]] of [[irreducible group representation|irreps]] some nontrivial $g$ will lie in the [[kernel]] of each of these. This means there exists a nontrivial irrep $\rho$ and nontrivial $g \in G$ with $\rho_{g}=I$ and correspondingly $\chi_{\rho}(g)= \dim \rho$, which appears in the [[character table]] of $G$ in a nontrivial row and column other than $[e]