 Solution

ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2017/2/b/i/solution

n_{7} ∣ 10

and

n_{7} \equiv 1 (mod 7)

. Among the divisors

1, 2, 5, 10

, only

1

satisfies the modular congruence. Hence

n_{7} = 1, K ◃ G, ∣ K ∣ = 7.

Uniqueness and the conjugacy assertion of the Sylow theorems make

K

 New to topics? Read the docs here!