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

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.