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Revision difference : $C_{mn}\cong C_m\times C_n$ when $m, n$ are relatively prime |
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Version 1 |
| We show that the cyclic group $C_{mn}$, where gcd$(m, n)=1$, is isomorphic to $C_m\times C_n$, where $C_m$ and $C_n$ are cyclic groups of orders $m$ and $n$, respectively. |
We show that the cyclic group $C_{mn}$, where gcd$(m, n)=1$, is isomorphic to $C_m\times C_n$, where $C_m$ and $C_n$ are cyclic groups of orders $m$ and $n$, respectively. |
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| 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 $0\leq x\leq m-1, 0\leq y\leq n-1$. |
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 $0\leq x\leq m-1, 0\leq y\leq n-1$. |
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| Next, we show that $C_m\times C_n$, a group, is cyclic. We do so by showing that it is generated by an element, namely $(x, y)$: |
Next, we show that $C_m\times C_n$, a group, 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 |
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 |
| \begin{eqnarray*} |
\begin{eqnarray*} |
| k &\equiv& i\;(mod\;m) \\ |
k &\equiv& i\;(mod\;m) \\ |
| k &\equiv& j\;(mod\;n). |
k &\equiv& j\;(mod\;n). |
| \end{eqnarray*} |
\end{eqnarray*} |
| 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. |
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. |
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| 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 $C_{mn}\cong C_m\times C_n$. |
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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