Exercise 2.1.11 (Some subgroups of $A\times B$)

Proof. Consider the canonical projections ${\pi }_1:A\times B\to A$ and ${\pi }_2:A\times \to B$ defined by ${\pi }_1((a,b))=a$ and ${\pi }_2((a,b))=b$. We know by Exercise 1.6.15 that ${\pi }_1$ and ${\pi }_2$ are homomorphisms.

(a)

Put $H=\{(a,1)\mid a\in A\}$. Then $$(a,b)\in H⟺a=1⟺(a,b)\in \ker <!--\; FUNCTION\; APPLICATION\; -->({\pi }_1).$$

Therefore $H=\{(a,1)\mid a\in A\}$ is a subgroup of $A\times B$(see Ex. 1.6.14) .

(b)

Similarly $$K=\{(1,b)\}\mid b\in B\}=\ker <!--\; FUNCTION\; APPLICATION\; -->({\pi }_2)$$

is a subgroup of $A\times B.$

(c)

Consider the map $\lambda :A\to A\times A$ defined by $\lambda (a)=(a,a)$. Then $\lambda$ is a homomorphism: for all $a\in A$ and for all $b\in A$, $$\lambda (a)\lambda (b)=(a,a)(b,b)=(\mathit{ab},\mathit{ab})=\lambda (\mathit{ab}).$$

Put $L=\{(a,a)\mid a\in A\}=\{(a,b)\in A\times A\mid a=b\}$.

If $(a,b)\in A\times B$, then $(a,b)\in L⟺a=b⟺(a,b)\in \lambda (A)$, so

$$L=\{(a,a)\mid a\in A\}=\lambda (A)$$

is a subgroup of $A\times A$ (see Ex. 1.6.13).