Exercise 3.3.4 ( $(A \times B)/(C \times D) \simeq (A/C) \times (B/D)$)

Proof.

Let $(a,b)\in A\times B$, and $(c,d)\in C\times D$. Since $C⊴A$ and $D⊴B$

so

Let ${\pi }_1:A\to A∕C$ and ${\pi }_2:B\to B∕D$ be the natural projections, i.e., ${\pi }_1(a)=\mathit{𝑎𝐶}$ for any $a\in A$ and ${\pi }_2(b)=\mathit{𝑏𝐷}$ for any $b\in B$.

Consider the map

Then

• $\phi$ is a homomorphism: If $(a,b)\in A\times B$ and $(a^{\prime },b^{\prime })\in A\times B$, then

  $$\begin{array}{llll}\hfill \phi ((a,b)(a^{\prime },b^{\prime }))& =\phi (a{a}^{\prime },b{b}^{\prime })=(a{a}^{\prime }C,b{b}^{\prime }D)& \hfill & \\ \hfill \phi (a,b)\phi (a^{\prime },b^{\prime })& =(\mathit{𝑎𝐶},\mathit{𝑏𝐷})(a^{\prime }C,b^{\prime }D)=(\mathit{𝑎𝐶}{a}^{\prime }C,\mathit{𝑏𝐷}{b}^{\prime }D)=(a{a}^{\prime }C,b{b}^{\prime }D),& \hfill & \end{array}$$

  so $\phi ((a,b)(a^{\prime },b^{\prime }))=\phi (a,b)\phi (c,d)$. Then $\phi$ is a homomorphism.

• $\phi$ is surjective. If $(\bar{a},\bar{b})\in (A∕C)\times (C∕D)$, then $\bar{a}=\mathit{𝑎𝐶}$ and $\bar{b}=\mathit{𝑏𝐷}$ for some $a\in A$ and some $b\in B$, because ${\pi }_1$ and ${\pi }_2$ are surjective. Therefore $(\bar{a},\bar{b})=({\pi }_1(a),{\pi }_2(b))=\phi (a,b)$, so $\phi$ is surjective, and

  $$\phi (A\times B)=(A∕C)\times (B∕D).$$

• $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=C\times D$: if $(a,b)\in A\times B$,

  $$\begin{array}{llll}\hfill (a,b)\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )& ⟺\phi (a,b)=(1_{A∕C},1_{B∕D})& \hfill & \\ \hfill & ⟺{\pi }_1(a)=1_{A∕B}\text{ and }{\pi }_2(b)=1_{C∕D}& \hfill & \\ \hfill & ⟺a\in \ker <!--\; FUNCTION\; APPLICATION\; -->({\pi }_1)\text{ and }b\in \ker <!--\; FUNCTION\; APPLICATION\; -->({\pi }_2)& \hfill & \\ \hfill & ⟺a\in C\text{ and }b\in D& \hfill & \\ \hfill & ⟺(a,b)\in C\times D,& \hfill & \end{array}$$

  so

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=C\times D.$$

By the First Isomorphism Theorem, $(A\times B)∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\simeq \phi (A\times B)$, so

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