The easiest way to build a new group from two old ones is to let them run side by side. This is also, as we will see, how every finite abelian group is built.
Definition. The direct product $G \times H$ is the set of pairs $(g, h)$ with the operation done coordinate by coordinate: $(g, h)(g', h') = (gg', hh')$. The identity is $(e, e)$, the inverse of $(g, h)$ is $(g^{-1}, h^{-1})$, and $|G \times H| = |G|\,|H|$. For additive groups we write $(a, b) + (a', b') = (a + a', b + b')$.
Why it works
Proof. $(g, h)^{m} = (g^{m}, h^{m})$ is the identity exactly when $g^{m} = e$ and $h^{m} = e$, that is, when $|g| \mid m$ and $|h| \mid m$ (order lemma). The least such $m$ is the lcm. $\square$
Theorem. $\mathbb{Z}_{m} \times \mathbb{Z}_{n}$ is cyclic, and then $\cong \mathbb{Z}_{mn}$, exactly when $\gcd(m, n) = 1$.
Proof. Every element $(a, b)$ has order dividing $\operatorname{lcm}(m, n)$, and $(1, 1)$ has order exactly $\operatorname{lcm}(m, n)$. The group has $mn$ elements, so it is cyclic exactly when $\operatorname{lcm}(m, n) = mn$, that is, when $\gcd(m, n) = 1$; then $(1, 1)$ generates it, and every cyclic group of order $mn$ is $\cong \mathbb{Z}_{mn}$. $\square$ This is the Chinese remainder theorem in group language.
Worked example: Order of an element
- $|4| = 12/\gcd(12, 4) = 3$ in $\mathbb{Z}_{12}$; $|6| = 15/\gcd(15, 6) = 5$ in $\mathbb{Z}_{15}$.
- $|(4, 6)| = \operatorname{lcm}(3, 5) = 15$.
Answer: $15$.
Worked example: Cyclic or not
- $\gcd(4, 6) = 2 \ne 1$, so it is not cyclic.
- The largest order is $\operatorname{lcm}(4, 6) = 12$, reached by $(1, 1)$, while the group has $24$ elements.
Answer: Not cyclic; largest order $12$.
Worked example: Counting elements of order 2
- Order dividing $2$ means $2(a, b) = (0, 0)$: $a \in \{0, 1\}$ and $b \in \{0, 2\}$, giving $4$ elements.
- Remove $(0, 0)$: $(1, 0)$, $(0, 2)$, $(1, 2)$.
Answer: $3$, while $\mathbb{Z}_{8}$ has only one. So $\mathbb{Z}_{2} \times \mathbb{Z}_{4} \not\cong \mathbb{Z}_{8}$.
Many familiar groups are products in disguise. $U(8) \cong \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ and $U(15) \cong \mathbb{Z}_{2} \times \mathbb{Z}_{4}$; in general, by the Chinese remainder theorem, $U(mn) \cong U(m) \times U(n)$ when $\gcd(m, n) = 1$.
Common mistake
Key idea