Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 3 / 11G / ii

The symmetric group $S_5$ has cardinality $5!=120=3\cdot 40$, so its Sylow subgroups for $p=3$ have cardinality $3$. One example isEvery nonidentity permutation of order of a group element $3$ in $S_5$ is a single permutation cycle of length $3$: two disjoint length-three permutation cycles would require six letters. Choose the three moved letters in $\left(\genfrac{}{}{0ex}{}{5}{3}\right)$ ways, then one of their two length-three permutation cycles, giving $20$ such elements. Each subgroup of cardinality $3$ has exactly two nonidentity elements, and distinct such subgroups have no nonidentity element in common. ConsequentlyThis also satisfies the numerical Sylow theorem restrictions $10\mid 40$ and $10\equiv 1(\mathrm{mod}3)$.