Let $\pi$ be a set of primes, and let $N$ have a single conjugacy class of $\pi$-complements and these are nilpotent. If $H^G / N$ is a $\pi$-group, then $H^G$ also has a single conjugacy class of nilpotent $\pi$-complements.
Why that?
2026-04-06 04:18:48.1775449128
Conclusion about conjugacy classes of $\pi$-complements
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