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

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.14

Solution: 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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