Exercise 1.1.29 ($A imes B$ is an abelian group if and only if both $A$ and $B$ are abelian)

Question

Prove that $A 	imes B$ is an abelian group if and only if both $A$ and $B$ are abelian.

Richard Ganaye · Accepted Answer

\begin{proof} 
If $(A, \star)$ and $(B, \diamond)$ are abelian groups, then for all $(a,b)$ and $(c,d)$ in $A	imes B$,
$$(a,b) (c,d) = (a \star c, b \diamond d) = (c \star a, d \diamond b) = (c,d) (a,b).$$
So $(A	imes B, \cdot)$ is an abelian group.

\bigskip conversely, suppose that $(A	imes B, \cdot)$ is an abelian group. Then $(a,b) (c,d) = (c,d) (a,b)$ for all $(a,b)$ and $(c,d)$ in $A	imes B$, so
 $$(a \star c, b \diamond d) = (c \star a, d \diamond b).$$
 Therefore $a \star c = c \star a$ for all $a,c \in A$, and $b \diamond d =  d \diamond b$ for all $b,d \in B$, so $(A, \star)$ and $(B, \diamond)$ are abelian groups.
\end{proof}

Exercise 1.1.29 ($A \times B$ is an abelian group if and only if both $A$ and $B$ are abelian)

Answers

Comments

Exercise 1.1.29 ($A \times B$ is an abelian group if and only if both $A$ and $B$ are abelian)

Answers

Comments

Add answer