Sylow I gives a <subgroup> of order equal to the full $p$-part of $|G|$. Sylow II says every $p$-subgroup lies in a conjugate of any Sylow $p$-subgroup; act by the <subgroup> on the Sylow cosets and use a fixed orbit because the number of cosets is prime to $p$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 2 / 9E / a / Solution

 Solution (source code)