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[parent] $C_{mn}\cong C_m\times C_n$ when $m, n$ are relatively prime (Proof)

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=\langle x\rangle$ and $ C_n=\langle y\rangle$. Then the external direct product $ C_m\times C_n$ consists of elements $ (x^i, y^j)$, where $ 0\leq i\leq m-1$ and $ 0\leq j\leq 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, \ldots, mn-1\}$. Such $ k$, if exists, would satisfy

$\displaystyle k$ $\displaystyle \equiv$ $\displaystyle i\;(mod\;m)$  
$\displaystyle k$ $\displaystyle \equiv$ $\displaystyle j\;(mod\;n).$  

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$.



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Cross-references: primality, Chinese remainder theorem, generates, generated by, cyclic, group, direct product, integer, positive, order, cyclic group, isomorphic, gcd

This is version 5 of $C_{mn}\cong C_m\times C_n$ when $m, n$ are relatively prime, born on 2008-04-16, modified 2008-04-17.
Object id is 10508, canonical name is C_mncongC_mtimesC_nWhenMNAreRelativelyPrime.
Accessed 260 times total.

Classification:
AMS MSC20A05 (Group theory and generalizations :: Foundations :: Axiomatics and elementary properties)

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