If you find any mistakes, please make a comment! Thank you.

A finite direct product of groups is a group under componentwise multiplication


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.

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu