= Solution
For a finite <group> of order $p^am$ 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 $56=7\cdot8$, 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 $8\cdot6=48$ 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 \b[a <group> of order 56 has a <normal subgroup> of order seven or of order eight].
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