Past exam of the short options course of the University of Oxford / 2017 / 2 / b / i

The Sylow theorems give $n_7\mid 10$ and $n_7\equiv 1(\mathrm{mod}7)$. Among the divisors $1,2,5,10$, only $1$ satisfies the modular congruence. HenceUniqueness and the conjugacy assertion of the Sylow theorems make $K$ a normal subgroup.