Past exam of the algebra course of the University of Oxford / 2008 / B / 8

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 givesA group of order $112=16\cdot 7$ has a Sylow two-subgroup of index seven. If the group were simple, the preceding result would give $112\mid 7!/2=2520$. This fails because $112$ is divisible by $16$ while $2520$ has only three factors of two. Thus 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 $8\cdot 6=48$ 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 no group of order $56$ is simple.

For order $105=3\cdot 5\cdot 7$, Sylow's congruence and divisibility conditions giveIf neither Sylow subgroup were a normal subgroup, there would be $21\cdot 4=84$ elements of order five and $15\cdot 6=90$ elements of order seven. Distinct subgroups of either prime order intersect only at the identity, and elements of the two orders are disjoint. This would give at least $84+90+1=175$ elements in a group of order $105$, a contradiction. Thus a subgroup of order five or seven is a normal subgroup, and the group is not simple.