A quotient by a product is isomorphic to the product of quotients

Let G = A_1 \times A_2 \times \cdots \times A_n and for each i let B_i \leq A_i be a normal subgroup. Prove that B_1 \times \cdots \times B_n \leq G is normal and that (A_1 \times \cdots \times A_n)/(B_1 \times \cdots \times B_n) \cong (A_1/B_1) \times \cdots \times (A_n/B_n).

We begin with some lemmas.

Lemma 1: Let \varphi_1 : G_1 \rightarrow H_1 and \varphi_2 : G_2 \rightarrow H_2 be group homomorphisms. Then \mathsf{ker}(\varphi_1 \times \varphi_2) = (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2). Proof: (\subseteq) Let (a,b) \in \mathsf{ker}(\varphi_1 \times \varphi_2). Then (1,1) = (\varphi_1 \times \varphi_2)(a,b) = (\varphi_1(a), \varphi_2(b)), hence \varphi_1(a) = 1 and \varphi_2(b) = 1. Thus a \in \mathsf{ker}\ \varphi_1 and b \in \mathsf{ker}\ \varphi_2, and (a,b) \in (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2). (\supseteq) If (a,b) \in (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2), then (\varphi_1 \times \varphi_2)(a,b) = (1,1); hence (a,b) \in \mathsf{ker}(\varphi_1 \times \varphi_2). \square

Lemma 2: Let A_1 and A_2 be groups with normal subgroups B_1 \leq A_1 and B_2 \leq A_2. Then B_1 \times B_2 is normal in A_1 \times A_2 and (A_1 \times A_2)/(B_1 \times B_2) \cong (A_1/B_1) \times (A_2/B_2). Proof: Let \pi_1 : A_1 \rightarrow A_1/B_1 and \pi_2 : A_2 \rightarrow A_2/B_2 denote the natural projections. These mappings are surjective, so that \pi_1 \times \pi_2 is a surjective homomorphism A_1 \times A_2 \rightarrow (A_1/B_1) \times (A_2/B_2). By Lemma 1, we have \mathsf{ker}(\pi_1 \times \pi_2) = B_1 \times B_2. The conclusion follows by the First Isomorphism Theorem. \square

The main result now follows by induction on n.

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