Sylow \(p\)-subgroups of order-56 groups (D&F 4.5.13)

Prove that a group of order 56 has a normal Sylow \(p\)-subgroup for some prime \(p\) dividing its order.

The primes diving 56 are 2 and 7, and a quick check shows that the possibilities for \(n_2\) and \(n_7\) are \(1,7\) and \(1,8\) respectively. Suppose that \(n_2 = 7\) and \(n_7 = 8\). All nonidentity elements in a Sylow 7-subgroup have order 7, and the intersection of 2 distinct Sylow 7-subgroups is trivial. Thus there are \(8 \cdot (7-1) = 48\) distinct elements of order 7. Since elements of a Sylow 2-subgroup cannot have order 7, they must be contained in the remaining 8 elements. But a Sylow 2-subgroup already has 8 elements, so there cannot possibly be 7 of them, since they would all be equal. We can thus conclude that either \(n_2\) or \(n_7\) (or both) is 1.