Constructive proof of $\forall n, m\in\mathbb N$ such that $\gcd (m,n)=1, $ the cyclic group $C_{nm}\cong C_n\times C_m$.

Question

Theorem: $\forall n, m\in\mathbb N$ such that $\gcd (m,n)=1, $ the cyclic group $C_{nm}\cong C_n\times C_m$.

This is a corollary of Chinese Remainder Theorem.

The theorem can be recast as saying that doing cyclic permutation on a list of all elements in the product is equivalent to doing so to each of the components. So, there has to be a more direct and intuitive approach. In other words, my goal is to find a constructive proof by explicitly constructing an isomorphism without using CRT (maybe by induction over $n$?), but I don't know where to start since I'm not familiar with permutations. Can anybody give me a hint?

The groups $C_4$ and $C_2 \oplus C_2$ are not isomorphic; the latter contains no element of (exactly) 4, for example. — anomaly, Oct 25 '19 at 19:06
You may find Sylow and Cauchy Theorem's to be helpful in finding elements of order p which divides one but not the other in creating the isomorphism — Andrew Shedlock, Oct 25 '19 at 19:15
Certainly this has been asked before, it's a classic abstract algebra homework question. — coco, Oct 25 '19 at 19:16

score 2 · Answer 1 · answered Oct 25 '19 at 19:25

2

Hint By the extended Euclidian Algorithm, there exists some $a,b$ such that $$am+bn=1$$

Show that $f(x)=(ax, bx)$ (or $x^a, x^b$ if you write the cyclic groups multiplicatively) is your isomorphism.

answered Oct 25 '19 at 19:25

N. S.

132,525

score 1 · Answer 2 · answered Oct 25 '19 at 19:22

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Hint.What is the order of $(g,h)$ when each of $g$ and $h$ generates the component it's in?

answered Oct 25 '19 at 19:22

Ethan Bolker

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pxc3110 · Answer 3 · 2019-10-25T20:37:42.983

0

Since cyclic groups are isomorphic to additive groups, set of all proofs of this theorem is "isomorphic", in a sense, to set of all proofs of CRT. Hence, find a constructive proof of CRT, which contains an isomorphism between $\mathbb Z_{nm}$ and $\mathbb Z_{m} \times \mathbb Z_{n}$. Replace every number by its corresponding cyclic permutation to obtain a constructive proof of this theorem.

edited Oct 25 '19 at 20:37

answered Oct 25 '19 at 20:24

pxc3110

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Constructive proof of $\forall n, m\in\mathbb N$ such that $\gcd (m,n)=1, $ the cyclic group $C_{nm}\cong C_n\times C_m$.

3 Answers3