Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 2 / 9E (Groups, Rings and Modules)

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 $p$.

Let an order-$p^{d-1}$ subgroup $H$ act on $G/H$. Since the number of cosets is divisible by $p$, orbit counting and Cauchy produce a fixed coset with stabiliser extending $H$ by index $p$. Induction yields subgroups of every $p$-power order. Applying the same orbit count to the action of $H$ on $G/H$ leaves fixed cosets indexed by $N(H)/H$, giving $|G:H|\equiv |N(H):H|(\mathrm{mod}p)$.

No. In an abelian group conjugacy is equality; for example distinct order-$p$ subgroups of $C_p\times C_p\times C_p$ are non-maximal and not conjugate.