 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/8/a/solution

Let

n = ord (a)

, which is finite. A group homomorphism preserves powers and the identity, so

f (a)^{n} = f (a^{n}) = f (e) = e^{'} .

r = ord (f (a))

, divide

n

r

n = q r + s

with

0 \leq s < r

. Then

e^{'} = f (a)^{n} = f (a)^{s}

, and minimality of

r

s = 0

. Thus

ord (f (a)) ∣ ord (a), ord (f (a)) \leq ord (a) .

f

is a group isomorphism, apply the same argument to its inverse:

ord (a) \leq ord (f (a))

. The two inequalities yield

ord (f (a)) = ord (a) .

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