----- > [!proposition] Proposition. ([[coprime characterization for when direct product is cyclic]]) > Let $m,n \in \mathbb{N}$. The [[direct product of groups|direct product]] of [[cyclic group]]s $C_{m} \times C_{n}$ is [[group isomorphism|isomorphic]] to $C_{mn}$ if and only if $m,n$ are [[relatively prime integers|coprime]]. ^4bba2c > [!proof]- Proof. ([[coprime characterization for when direct product is cyclic]]) \ $\to$. Assume $m,n$ are [[relatively prime integers|coprime]], so that [[greatest common divisor]]$(m,n)=1$. Define the map $\begin{align} \psi:\mathbb{Z}_{mn} \to & \mathbb{Z}_{m} \times \mathbb{Z}_{n} \\ \psi(\ell) := & (\ell \text{ mod }m, \ell \text{ mod }n). \end{align}$ \ Since $m$, $n$ are [[relatively prime integers|coprime]], the [[Chinese remainder theorem]] tells us that the system $\begin{align} x \text{ mod } m = a \\ x \text{ mod }n = b \end{align}$ has a unique solution (mod $mn$). Thus, if $(a,b) \in C_{m} \times C_{n}$ then there exists $x \in [mn]$ such that $(a,b)=(x \text{ mod }m, x \text{ mod }n)$. Hence $\psi$ is [[surjection|surjective]]. The uniqueness of $x$ also guarantees $\psi$ is [[injection|injective]]. Thus all we need to show is $\psi \in \hom(\mathbb{Z}_{mn}, \mathbb{Z}_{m} \times \mathbb{Z}_{n})$. We have$\begin{align}\psi(\ell_{1}+\ell_{2})= & ((\ell_{1}+\ell_{2}) \text{ mod } m, (\ell_{1} + \ell _{2}) \text{ mod } n) \\ = & (\ell_{1} \text{ mod } m + \ell _{2} \text{ mod m}, \ell_{1} \text{ mod }n + \ell_{2} \text{ mod }n ) \\ = & (\ell_{1} \text{ mod }m , \ell_{1} \text{ mod n}) + (\ell_{2} \text{ mod }m + \ell_{2} \text{ mod }n) \\ = & \psi(\ell_{1}) + \psi(\ell_{2}),\end{align}$ as required. $\leftarrow$. Assume $m,n$ are not [[relatively prime integers|coprime]], $\text{gcd}(m,n)=k>1$. Let $G=C_{m} \times C_{n}$. Then for any $(a,b) \in G$, we have $(a,b)^{\text{lcm}(m,n)}=e$. But [[least common multiple]]$(m,n)=\frac{mn}{k}$ by [[product of integers is product of gcd and lcm]], and since $\frac{mn}{k}<mn$, we conclude that no element of $G$ can have order $mn$. So $G$ is not a [[cyclic group]]. ^5a3562 -----. #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```