Exercise 1.1.30 (Order of $(a,b)$ in $A imes B$)

Question

Prove that the elements $(a,1)$ and $(1,b)$ of $A 	imes B$ commute and deduce that the order of $(a,b)$ is the least common multiple of $|a|$ and $|b|$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof}
First $(a,1) (1,b) = (a \star 1, 1 \diamond b) = (a,b)$ and $(1,b)(a,1) = (1 \star a, b \diamond 1) = (a,b)$, so the elements $(a,1)$ and $(1,b)$ of $A 	imes B$ commute.

Since these two elements commute, by Exercise 24, $(a,b)^n = (a,1)^n (1,b)^n =(a^n, 1) (1,b^n) =(a^n, b^n)$ for all $n \in \Z$. Then for all integers $n$,
\begin{align*}
(a,b)^n = 1_{A	imes B} &\iff (a^n, b^n) = (1,1)\
& \iff
\begin{cases}
a^n =1 \
b^n = 1 
\end{cases}\
&\iff 
\begin{cases}
n \mid |a|\
n \mid |b|
\end{cases}\
&\iff n \mid \mathrm{l.c.m.}(|a|,|b|)).
\end{align*}
Therefore
$$|(a,b)| = \mathrm{l.c.m.}(|a|,|b|).$$
\end{proof}

Exercise 1.1.30 (Order of $(a,b)$ in $A \times B$)

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Exercise 1.1.30 (Order of $(a,b)$ in $A \times B$)

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