 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-i/5/b/iv/solution

T

is a surjective group homomorphism, then

G = T (G)

, which is an abelian group by the preceding part.

T

is an injective group homomorphism, then for all

a, b \in G

T (ab) = T (a) T (b) = T (b) T (a) = T (ba),

because

T (a), T (b)

T (G)

. Injectivity gives

ab = ba

. Hence

G is abelian in either case

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