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

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., $(a_1, b_1) \cdot (a_2, b_2) = (a_1 a_2, b_1 b_2)$ .

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

Let $a_i \in A$ and $b_i \in B$ for $i = 1,2,3$ . Then we have

$(a_1, b_1) \cdot \left[ (a_2, b_2) \cdot (a_3, b_3) \right]$ = $(a_1, b_1) \cdot (a_2 a_3, b_2 b_3)$

= $(a_1 (a_2 a_3), b_1 (b_2 b_3))$

= $((a_1 a_2) a_3, (b_1 b_2) b_3)$

= $(a_1 a_2, b_1 b_2) \cdot (a_3, b_3)$

= $\left[ (a_1, b_1) \cdot (a_2, b_2) \right] \cdot (a_3, b_3)$

So componentwise multiplication is indeed associative.
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 $A \times B$ .
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. $\blacksquare$

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