 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/9/a/solution

f : G \to G^{'}

satisfies

f (g h) = f (g) f (h)

for all

g, h \in G

. It consequently sends the identity element to the identity and inverses to inverses. Its group homomorphism kernel is

ker f = {g \in G : f (g) = e_{G^{'}}} .

H

G

is a subgroup invariant under conjugation:

g H g^{- 1} = H for every g \in G .

Equivalently,

g H = H g

for every

g

, so left and right cosets coincide.

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