# How to find normal subgroups from a character table?

I know that normal subgroups are the union of some conjugacy classes

Conjugacy classes are represented by the the columns in a matrix

How could we use character values in the table to determine normal subgroups?

First fact : $$N$$ is a normal subgroup of a finite group $$G$$ if and only if there exists a character $$\chi$$ of $$G$$ such that $$N = \ker \chi := \{g \in G | \chi(g)=\chi(1)\}$$. Indeed, if $$N$$ is normal then $$G$$ acts on the complex algebra $$\mathbf{C}[G/N] = \displaystyle \bigoplus_{gN \in G/N} \mathbf{C} e_{gN}$$ by $$h \cdot e_{gN}=e_{hgN}$$. This is a linear representation of $$G$$ (coming from the regular representation of $$G/N$$). Let $$\chi$$ be its character. It is easy to check that $$\chi(h) = 0$$ if $$h \notin N$$ and $$\chi(h) = \mathrm{Card}(G/N) = \chi(1)$$ if $$h \in N$$. So $$N = \ker \chi$$. Conversely, using the fact that a character is constant on every conjugacy class, any subgroup of the form $$\ker \chi$$ is normal.
Second fact : if $$\rho : G \to \mathrm{GL}(V)$$ is the representation associated to the character $$\chi$$ then $$\ker \rho = \ker \chi$$. The inclusion $$\subseteq$$ is trivial. Conversely, assume $$\chi(g) = \chi(1) = \dim V$$. Since the eigenvalues of $$\rho(g)$$ are roots of $$1$$ and $$\chi(g)$$ is the sum of the eigenvalues (with multiplicities), these eigenvalues are forced to be all equal to $$1$$. So $$\rho(g) = \mathrm{id}_V$$, that is to say $$g \in \ker \rho$$.
Third fact : if $$\chi = \displaystyle \sum_{i=1}^r n_i \chi_i$$ (where the $$\chi_i$$ are pairwise distinct irreducible characters and $$n_i \geq 1$$) then $$\ker \chi = \displaystyle \bigcap_{i=1}^r \ker \chi_i$$. Writing $$\rho, \rho_1,\ldots,\rho_r$$ for the corresponding representations, $$\rho$$ is the direct sum of copies of $$\rho_1,\ldots,\rho_r$$ so $$\ker \rho = \displaystyle \bigcap_{i=1}^r \ker \rho_i$$. Then apply the second fact.
Conclusion : with your character table, you can read the subgroups $$N_i:=\ker \chi_i$$. Then the normal subgroups of $$G$$ are exactly the intersections of some $$N_i$$.