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Is it possible to determine if a group is simple by only looking at the character table?

My idea was to find the order of the group by summing the squares of the first column and then using Burnside's theorem. But of course, this works for a very special case. Alternately, I know that given a normal subgroup $N$ in $G$, an irreducible representation of $G/N$ can be extended to an irreducible representation of $G$. Perhaps this may be of use, since normal subgroups are union of conjugacy classes?

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From Wikipedia: "All normal subgroups of $G$ (and thus whether or not $G$ is simple) can be recognised from its character table. The kernel of a character $\chi$ is the set of elements $g \in G$ for which $\chi(g) = \chi(1)$; this is a normal subgroup of $G$. Each normal subgroup of $G$ is the intersection of the kernels of some of the irreducible characters of $G$."

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    $\begingroup$ One question: how do I show they form a group? I'm stuck in the part where $\chi(g)=\chi(g')=\chi(1) \implies \chi(gg')=\chi(1)$ $\endgroup$
    – kleinbottle
    Apr 28, 2022 at 6:18
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    $\begingroup$ I think the facts that kernels of homomorphisms are subgroups and intersections of subgroups is a subgroup are enough. $\endgroup$
    – Bailey
    Apr 28, 2022 at 6:25

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