 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/9/b/solution

ϕ : H \to G

ϕ (h_{1} h_{2}) = ϕ (h_{1}) ϕ (h_{2}) for all h_{1}, h_{2} \in H .

It necessarily sends

e_{H}

e_{G}

and sends inverses to inverses, by applying this relation to

e_{H}^{2}

and

h h^{- 1}

ker ϕ = {h \in H : ϕ (h) = e_{G}} .

The kernel is a normal subgroup of the domain: if

k \in ker ϕ

and

h \in H

, then

ϕ (hk h^{- 1}) = ϕ (h) e_{G} ϕ (h)^{- 1} = e_{G}

. The kernel’s identity condition refers to the identity in the target group.

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