Algebra Solutions: Compute the order of an element in a direct product of groups

Let $A$ and $B$ be groups and let $a \in A$ and $b \in B$ have finite order. Prove that the elements $(a,1)$ and $(1,b)$ commute in $A \times B$ and deduce that the order of $(a,b)$ is the least common multiple of $|a|$ and $|b|$ .

First we prove a few lemmas.
Lemma 1. If $a \in G$ has finite order, say $|a| = n$ , and $a^m = 1$ , then $n|m$ . Proof: By the division algorithm, there exist unique $(q,r)$ such that $0 \leq r < |n|$ and $m = qn + r$ . Then we have $1 = a^m = a^{qn + r} = (a^n)^q a^r = a^r$ . But $n$ is the least positive integer such that $a^n = 1$ ; hence $r = 0$ and we have $m = qn$ . So $n|m$ . $\square$
Lemma 2. If $a, b \in G$ such that $ab = ba$ and $\{ a^n \ |\ n \in \mathbb{Z} \} \cap \{ b^n \ |\ n \in \mathbb{Z} \} = \{ 1 \}$ (i.e., $a$ and $b$ have only trivial powers in common) then $|ab| = \mathsf{lcm}(|a|, |b|)$ . Proof: Let $|a| = n_1$ , $|b| = n_2$ , and $|ab| = m$ , and let $c = \mathsf{lcm}(n_1, n_2)$ . Then using a previous theorem and Lemma 1, we have $(ab)^c = a^c b^c = 1$ so that $m|c$ . Now since $(ab)^m = a^m b^m = 1$ , we have $a^m = b^{-m}$ . Since $a$ and $b$ have only trivial powers in common, $a^m = 1$ and $b^m = 1$ , so that $n_1 | m$ and $n_2 | m$ . By the definition of least common multiple, $c|m$ . So $c = m$ . $\square$
Lemma 3. For all $n \in \mathbb{Z}$ , $(a,b)^n = (a^n, b^n)$ . Proof: We have $(a,b)^0 = 1 = (1,1) = (a^0, b^0)$ . We prove the lemma for $n > 0$ by induction. For the base case $n = 1$ we have $(a,b)^1 = (a,b) = (a^1, b^1)$ . Now suppose the lemma holds for som $n \geq 1$ ; then $(a,b)^{n+1} = (a,b)^n (a,b) = (a^n,b^n) (a,b) = (a^{n+1}, b^{n+1})$ . Thus by induction the lemma holds for all positive $n$ . For $n < 0$ , we have $(a,b)^n = ((a,b)^{-n})^{-1} = (a^{-n}, b^{-n})^{-1} = (a^n,b^n)$ , and the lemma holds. $\square$
Now for the main result it suffices to show that under the hypotheses, $(a,1)$ and $(1,b)$ have only the trivial power in common. This follows because $(a,1)^n = (a^n,1)$ and $(1,b)^m = (1,b^m)$ for all $n$ and $m$ , so that $(a,1)^n = (1,b)^m$ implies in particular that $a^n = 1$ and so $(a,1)^n = (1,1)$ . The conclusion follows from Lemma 2. $\blacksquare$

Algebra Solutions

Compute the order of an element in a direct product of groups

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