The normal subgroups of a finite group are exactly the intersections of kernels of irreducible complex characters

Assume $N⊴G$. Let $\bar{\rho }:G\mathrm{/}N\to \mathrm{GL}(\mathbb{C}[G\mathrm{/}N])$ be the regular representation of the quotient. Since $\bar{\rho }$ is faithful by [F4], its kernel in $G\mathrm{/}N$ is the trivial subgroup $\{N\}$.

Conversely, if $N={\bigcap }_j\ker {\chi }_j$ for irreducible characters ${\chi }_j$ of $G$, then by [F3] each $\ker {\chi }_j$ is the kernel of a group homomorphism, hence normal by [A2], and the intersection of normal subgroups is again normal by [A2]. Thus $N⊴G$.

Since $\mathrm{char}\mathbb{C}=0$ does not divide $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }$, [F1] decomposes $\mathbb{C}[G\mathrm{/}N]$ as a direct sum of irreducible representations $U_1,\dots ,U_m$ of $G\mathrm{/}N$; by [A1] the trivial kernel of step 1.1 is the intersection of their kernels, so ${\bigcap }_j\ker (\bar{\rho }{\mathrm{\mid }}_{U_j})=\{N\}$.

For each $j$, let $V_j$ be the inflation of $U_j$ to $G$. By [F2], each $V_j$ is irreducible as a representation of $G$, and $\ker {\rho }_{V_j}={\pi }^{-1}(\ker (\bar{\rho }{\mathrm{\mid }}_{U_j}))$, which contains $N$. By [F3] each $\ker {\rho }_{V_j}$ equals the kernel of the corresponding irreducible character ${\chi }_j$ of $G$.

By [A1] and steps 2.1 and 3.1, ${\bigcap }_j\ker {\chi }_j={\bigcap }_j{\pi }^{-1}(\ker (\bar{\rho }{\mathrm{\mid }}_{U_j}))={\pi }^{-1}(\{N\})=N$. Hence a normal subgroup is an intersection of kernels of irreducible characters.

Steps 4.1 and 1.2 prove the two implications, hence the equivalence.