**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.1 Exercise 2.1.11**

Let $A$ and $B$ be groups. Prove that the following sets are subgroups of $A \times B$.

(1) $A \times 1$,

(2) $1 \times B$,

(3) If $B = A$, the set $\Delta(A) = \{ (a,a) \ |\ a \in A \}$.

Solution:

(1) $A \times 1$ is not empty since $(1,1) \in A \times 1$. Now suppose $(a_1,1)$, $(a_2,1) \in A \times 1$. Then $a_1a_2^{-1} \in A$, so that $$(a_1,1) (a_2,1)^{-1} = (a_1a_2^{-1},1) \in A \times 1.$$ By the subgroup criterion, $A \times 1 \leq A \times B$.

(2) $1 \times B$ is not empty since $(1,1) \in 1 \times B$. Now suppose $(1,b_1)$, $(1,b_2) \in 1 \times B$. Then $b_1b_2^{-1} \in B$, so that $$(1,b_1) (1,b_2)^{-1} = (1,b_1b_2^{-1}) \in 1 \times B.$$ By the subgroup criterion, $1 \times B \leq A \times B$.

(3) $\Delta A$ is not empty since $(1,1) \in \Delta A$. Now suppose $(a,a)$, $(b,b) \in \Delta A$; then $$(a,a)(b,b)^{-1} = (ab^{-1},ab^{-1}) \in \Delta A.$$ By the subgroup criterion, $\Delta A \leq A \times A$.