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Will Sawin
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For nilpotency, you can deduce the character table of $G/Z$ from the character table of $Z$$G$. First, determine $Z$. Second, throw out all the representations where $Z$ is not in the kernel. Third, merge the conjugacy classes which have the same trace in every representation. (This works because the irreducible representations of $G/Z$ are exactly the irreducible representations of $G$ with kernel containing $Z$, and because the inverse images of the conjugacy classes of $G/Z$ in $G$ are unions of conjugacy classes of $G$.) Then iterate.

For nilpotency, you can deduce the character table of $G/Z$ from the character table of $Z$. First, determine $Z$. Second, throw out all the representations where $Z$ is not in the kernel. Third, merge the conjugacy classes which have the same trace in every representation. (This works because the irreducible representations of $G/Z$ are exactly the irreducible representations of $G$ with kernel containing $Z$, and because the inverse images of the conjugacy classes of $G/Z$ in $G$ are unions of conjugacy classes of $G$.) Then iterate.

For nilpotency, you can deduce the character table of $G/Z$ from the character table of $G$. First, determine $Z$. Second, throw out all the representations where $Z$ is not in the kernel. Third, merge the conjugacy classes which have the same trace in every representation. (This works because the irreducible representations of $G/Z$ are exactly the irreducible representations of $G$ with kernel containing $Z$, and because the inverse images of the conjugacy classes of $G/Z$ in $G$ are unions of conjugacy classes of $G$.) Then iterate.

Source Link
Will Sawin
  • 148.4k
  • 9
  • 324
  • 563

For nilpotency, you can deduce the character table of $G/Z$ from the character table of $Z$. First, determine $Z$. Second, throw out all the representations where $Z$ is not in the kernel. Third, merge the conjugacy classes which have the same trace in every representation. (This works because the irreducible representations of $G/Z$ are exactly the irreducible representations of $G$ with kernel containing $Z$, and because the inverse images of the conjugacy classes of $G/Z$ in $G$ are unions of conjugacy classes of $G$.) Then iterate.