Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1 / 2F (Groups, Rings and Modules)

The normalizer is $N_G(H)=\{g\in G:gH{g}^{-1}=H\}$. It is a subgroup: the identity normalizes $H$, products of normalizing elements normalize it, and conjugating the equality by $g^{-1}$ proves inverse closure. The mapis well-defined and onto. Two images are equal exactly when $g_2^{-1}{g}_1\in N_G(H)$, exactly the condition that their cosets coincide. Thus the number of conjugates is $\boxed{[G:N_G(H)]}$.

By the permitted conjugacy assertion for Sylow subgroups, the number $n_p$ of Sylow $p$-subgroups equals $[G:N_G(P)]$ for one such subgroup $P$. Lagrange's theorem makes this divide $|G|$, and $P\subseteq N_G(P)$ also makes it divide $|G|/|P|$.

For order $72=8\cdot 9$, a Sylow $3$-subgroup has order nine, so $n_3$ divides eight. To obtain the required congruence without assuming an additional Sylow counting theorem, let $P$ act by conjugation on the set of Sylow $3$-subgroups. Its orbit sizes are powers of three. A fixed subgroup $Q$ is normalized by $P$, so $PQ$ is a subgroup and its order $|P||Q|/|P\cap Q|$ is a power of three. Since nine is the largest power of three dividing $72$, this forces $PQ=P=Q$. There is exactly one fixed point, namely $P$ itself. All other orbits have sizes divisible by three, giving $n_3\equiv 1(\mathrm{mod}3)$. Among $1,2,4,8$, only one and four satisfy this. Therefore $\boxed{n_3\in \{1,4\}}$.