Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/b/b2/c/solution

Let . The map sending to is well defined: multiplying on the right by an element commuting with leaves its conjugate unchanged. It is onto by the orbit definition. If , then commutes with , so , proving injectivity.
Every coset has elements, so . Therefore
This is the orbit-stabilizer theorem specialized to conjugation.

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