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## In a finite group, conjugation permutes sub-Sylow subgroups

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.5 Exercise 4.5.2

Solution: Note that $gQg^{-1} \leq gHg^{-1}$. Moreover, because conjugation by g is an automorphism, $|gQg^{-1}| = |Q|$ and $|gHg^{-1}| = |H|$. Hence $$[gHg^{-1} : gQg^{-1}] = |gHg^{-1}|/|gQg^{-1}| = |H|/|Q| = [H : Q]$$ is not divisible by $p$, so that $gQg^{-1}$ is a Sylow $p$-subgroup of $gHg^{-1}$.