**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.28**

Let $A$ and $B$ be groups. Verify that $A \times B$ is a group under componentwise multiplication; i.e., $$(a_1, b_1) \cdot (a_2, b_2) = (a_1 a_2, b_1 b_2).$$

Solution: We need to verify the three group axioms: associativity, identity, and inverses.

(1) Let $a_i \in A$ and $b_i \in B$ for $i = 1,2,3$. Then we have \begin{align*} (a_1, b_1) \cdot \left[ (a_2, b_2) \cdot (a_3, b_3) \right] = & \ (a_1, b_1) \cdot (a_2 a_3, b_2 b_3)\\ =&\ (a_1 (a_2 a_3), b_1 (b_2 b_3))\\ = &\ ((a_1 a_2) a_3, (b_1 b_2) b_3)\\ = &\ (a_1 a_2, b_1 b_2) \cdot (a_3, b_3)\\ = &\ \left[ (a_1, b_1) \cdot (a_2, b_2) \right] \cdot (a_3, b_3) \end{align*} So componentwise multiplication is indeed associative.

(2) Note that for all $a \in A$ and $b \in B$, $$(a,b) \cdot (1_A, 1_B) = (a \cdot 1_A, b \cdot 1_B) = (a,b).$$ Similarly, $(1_A,1_B) \cdot (a,b) = (a,b)$. So $(1_A, 1_B)$ is an identity element in $A \times B$.

(3) Let $(a,b) \in A \times B$. Then $a^{-1} \in A$ and $b^{-1} \in B$, so that $(a^{-1}, b^{-1}) \in A \times B$. Moreover we have $$(a,b) \cdot (a^{-1}, b^{-1}) = (aa^{-1}, bb^{-1}) = (1_A, 1_B).$$ Thus every element of $A \times B$ has an inverse.

So $A \times B$ is indeed a group under componentwise multiplication.