Cmn ∼
= Cm × Cn when m, n are relatively
prime∗
yesitis†
2013-03-22 0:53:27
We show that Cmn , gcd(m, n) = 1, is isomorphic to Cm × Cn , where Cr
denotes the cyclic group of order r for any positive integer r.
Let Cm = hxi and Cn = hyi. Then the external direct product Cm × Cn
consists of elements (xi , y j ), where 0 ≤ i ≤ m − 1 and 0 ≤ j ≤ n − 1.
Next, we show that the group Cm × Cn is cyclic. We do so by showing
that it is generated by an element, namely (x, y): if (x, y) generates Cm × Cn ,
then for each (xi , y j ) ∈ Cm × Cn , we must have (xi , y j ) = (x, y)k for some
k ∈ {0, 1, 2, . . . , mn − 1}. Such k, if exists, would satisfy
k
≡ i (mod m)
k
≡ 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,
Cm × Cn is generated by (x, y), so it is cyclic.
The order of Cm × Cn is mn, so is the order of Cmn . Since cyclic groups of
the same order are isomorphic, we finally have Cmn ∼
= Cm × Cn .
∗ hCmncongCmtimesCnWhenMNAreRelativelyPrimei
created: h2013-03-2i by: hyesitisi
version: h40508i Privacy setting: h1i hProofi h20A05i
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