 Solution

For a finite group of order $p^{a}m$ with $p$ prime and $p\nmid m$, the Sylow theorems assert that a subgroup of order $p^a$ exists; every subgroup whose order is a power of $p$ is contained in one such Sylow subgroup; all Sylow subgroups for that prime are conjugate; and their number $n_p$ divides $m$ and is congruent to one modulo $p$. A unique Sylow subgroup is a normal subgroup, since conjugation cannot move it to another one.

Here $56=7\cdot 8$, so $n_7$ divides eight and is congruent to one modulo seven. Hence $n_7=1$ or $8$. In the first case the Sylow subgroup of order seven is normal. In the second case, any two distinct order-seven subgroups intersect only at the identity, because a nontrivial intersection would generate both prime-order subgroups. Their nonidentity elements therefore account for $8\cdot 6=48$ elements.

Exactly eight elements of the group remain. Every Sylow subgroup of order eight consists entirely of these remaining elements: a nonidentity element in it has order a power of two and cannot be one of the order-seven elements already counted. Each such Sylow subgroup must consequently equal the remaining eight-element set. It is unique and hence normal. Thus a group of order 56 has a normal subgroup of order seven or of order eight.