# ${C}_{mn}\cong {C}_{m}\times {C}_{n}$ when $m,n$ are relatively prime

We show that ${C}_{mn}$, gcd$(m,n)=1$, is isomorphic^{} to ${C}_{m}\times {C}_{n}$, where ${C}_{r}$ denotes the cyclic group^{} of order $r$ for any positive integer $r$.

Let ${C}_{m}=\u27e8x\u27e9$ and ${C}_{n}=\u27e8y\u27e9$. Then the external direct product ${C}_{m}\times {C}_{n}$ consists of elements $({x}^{i},{y}^{j})$, where $0\le i\le m-1$ and $0\le j\le n-1$.

Next, we show that the group ${C}_{m}\times {C}_{n}$ is cyclic. We do so by showing that it is generated by an element, namely $(x,y)$: if $(x,y)$ generates ${C}_{m}\times {C}_{n}$, then for each $({x}^{i},{y}^{j})\in {C}_{m}\times {C}_{n}$, we must have $({x}^{i},{y}^{j})={(x,y)}^{k}$ for some $k\in \{0,1,2,\mathrm{\dots},mn-1\}$. Such $k$, if exists, would satisfy

$k$ | $\equiv $ | $i(modm)$ | ||

$k$ | $\equiv $ | $j(modn).$ |

Indeed, by the Chinese Remainder Theorem^{}, such $k$ exists and is unique modulo $mn$. (Here is where the relative primality of $m,n$ comes into play.) Thus, ${C}_{m}\times {C}_{n}$ is generated by $(x,y)$, so it is cyclic.

The order of ${C}_{m}\times {C}_{n}$ is $mn$, so is the order of ${C}_{mn}$. Since cyclic groups of the same order are isomorphic, we finally have ${C}_{mn}\cong {C}_{m}\times {C}_{n}$.

Title | ${C}_{mn}\cong {C}_{m}\times {C}_{n}$ when $m,n$ are relatively prime |
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Canonical name | CmncongCmtimesCnWhenMNAreRelativelyPrime |

Date of creation | 2013-03-22 17:59:46 |

Last modified on | 2013-03-22 17:59:46 |

Owner | yesitis (13730) |

Last modified by | yesitis (13730) |

Numerical id | 8 |

Author | yesitis (13730) |

Entry type | Proof |

Classification | msc 20A05 |