**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.30**

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 $|b|$.

Solution: 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 $n|m$.

**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 Exercise 24 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 $c = m$.

**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 some $n \geq 1$; 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.

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.