= Solution
For the preliminary index assertion, the <coset action> gives a homomorphism $G\to S_n$. Its <kernel> is a <normal subgroup> and cannot be all of $G$, because the action is transitive on $n>1$ points. Simplicity makes the <kernel> trivial. The <sign homomorphism> of this faithful action must be trivial: otherwise simplicity would identify $G$ with $C_2$, impossible for a transitive action of degree at least three. Thus $G$ embeds in $A_n$, and Lagrange gives
$$
\boxed{|G|\mid n!/2.}
$$
A group of order $112=16\cdot7$ has a Sylow two-<subgroup> of index seven. If the group were simple, the preceding result would give $112\mid7!/2=2520$. This fails because $112$ is divisible by $16$ while $2520$ has only three factors of two. Thus \b[there is no <simple group> of order $112$].
For order $56$, Sylow gives $n_7\in\{1,8\}$. If it is one, the order-seven Sylow is a <normal subgroup>. Otherwise its eight Sylows of order seven intersect only at the identity, giving $8\cdot6=48$ nonidentity elements of order seven. Only eight elements remain including the identity. Every Sylow two-<subgroup> has order eight and consists entirely of these remaining elements. Hence all Sylow two-subgroups coincide and the order-eight <subgroup> is a <normal subgroup>. Either case gives a nontrivial proper <normal subgroup>, so \b[no group of order $56$ is simple].
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