Simple Group Kernel 92244D

1. **Problem Statement:** Prove that a finite group $G$ is simple if and only if the kernel of every non-principal irreducible character $\chi$ of $G$ is trivial, i.e., $\ker(\chi) = \{1_G\}$. 2. **Recall definitions and facts:** - A group $G$ is *simple* if it has no nontrivial normal subgroups. - For an irreducible character $\chi$ of $G$, the kernel $\ker(\chi) = \{g \in G : \chi(g) = \chi(1)\}$ is a normal subgroup of $G$. - The *principal character* $\chi_0$ is the trivial character with $\chi_0(g) = 1$ for all $g \in G$. 3. **Direction 1: If $G$ is simple, then $\ker(\chi) = \{1_G\}$ for all non-principal $\chi$:** - Since $G$ is simple, the only normal subgroups are $\{1_G\}$ and $G$ itself. - For any non-principal irreducible character $\chi$, $\ker(\chi)$ is a normal subgroup. - If $\ker(\chi) = G$, then $\chi$ is the principal character, contradicting the assumption. - Therefore, $\ker(\chi)$ must be $\{1_G\}$. 4. **Direction 2: If $\ker(\chi) = \{1_G\}$ for all non-principal $\chi$, then $G$ is simple:** - Suppose $N$ is a nontrivial normal subgroup of $G$. - Consider the quotient group $G/N$ and its irreducible characters. - By lifting characters from $G/N$ to $G$, there exists a non-principal irreducible character $\chi$ of $G$ with $N \subseteq \ker(\chi)$. - Since $N$ is nontrivial, $\ker(\chi) \neq \{1_G\}$, contradicting the assumption. - Hence, no nontrivial normal subgroup exists, so $G$ is simple. 5. **Summary:** - The kernel of any irreducible character is a normal subgroup. - The simplicity of $G$ forces these kernels to be trivial for all non-principal characters. - Conversely, trivial kernels for all non-principal irreducible characters imply no nontrivial normal subgroups. **Final conclusion:** $$\boxed{\text{$G$ is simple} \iff \ker(\chi) = \{1_G\} \text{ for all non-principal } \chi \in \mathrm{Irr}(G)}.$$