 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/9/a/solution

K = {g \in G : Φ (g) = e_{G^{'}}} .

The homomorphism sends identity to identity and inverses to inverses. Thus

e_{G} \in K

, and if

k_{1}, k_{2} \in K

, then

Φ (k_{1} k_{2}^{- 1}) = e_{G^{'}}

, so

K

For any

g \in G

and

k \in K

Φ (g k g^{- 1}) = Φ (g) Φ (k) Φ (g)^{- 1} = Φ (g) e_{G^{'}} Φ (g)^{- 1} = e_{G^{'}} .

Hence

g K g^{- 1} \subseteq K

. Applying the same inclusion with

g^{- 1}

gives equality, proving

K ⊴ G .

This normality argument does not require either group to be finite.

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