 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/8/f/solution

Closure of the componentwise operation follows from closure in each factor. For three pairs, associativity holds coordinatewise:

((g_{1}, g_{2}) (h_{1}, h_{2})) (k_{1}, k_{2}) = ((g_{1} h_{1}) k_{1}, (g_{2} h_{2}) k_{2}) = (g_{1} (h_{1} k_{1}), g_{2} (h_{2} k_{2})) .

(e_{1}, e_{2})

(g_{1}, g_{2})

(g_{1}^{- 1}, g_{2}^{- 1})

∣ G_{1} \times G_{2} ∣ = ∣ G_{1} ∣ ∣ G_{2} ∣.

N = G_{1} \times {e_{2}}

contains the identity and is closed under products and inverses, so it is a subgroup. For any

x = (a, b)

in the product and

(g, e_{2}) \in N

x (g, e_{2}) x^{- 1} = (a g a^{- 1}, b e_{2} b^{- 1}) = (a g a^{- 1}, e_{2}) \in N .

x^{- 1}

gives the reverse inclusion, and hence

x N x^{- 1} = N, N ⊴ G_{1} \times G_{2} .

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