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In general, products of cyclic groups are not cyclic. If $C_n$ is the cyclic group of order $n$, $C_{22} \times C_{33}$ is not cyclic. But is there an easy way to tell when such a product will be cyclic?

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marked as duplicate by YACP, Dennis Gulko, Davide Giraudo, rschwieb, Marc van Leeuwen Mar 18 '13 at 13:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
See also math.stackexchange.com/questions/272371/… –  user26857 Mar 18 '13 at 8:53

2 Answers 2

up vote 9 down vote accepted

Yes, the direct product of two cyclic groups, $C_m \times C_n$ is cyclic if and only if the $$\gcd(m, n) = 1$$ (where $m$ is the order of one of the cycle groups, and $n$ the order of the other.)

Note, we must have a Direct product of order $mn$.

  • If the $\gcd(m, n)\neq 1$, then the largest order of any element in the direct product is $\text{lcm}\,(m, n)\lt mn$, where $mn$ is the order of the Direct Product, and hence no element in the direct product could generate the entire direct product. And so, the direct product would not be a cyclic group.

  • When $\gcd(m,n) = 1$, then it follows that the largest order of any element in the direct product is of order $\text{lcm}\,(m, n) = mn$, and so there exists an element of order $mn \in C_m \times C_n$. Since we know that the order of $C_m\times C_n$ is $\,mn\,$ then there exists an element in $C_m\times C_n\,$ which generates the entire direct product. Hence, the direct product must by cyclic.

If $\gcd(m, n) = 1$, we have $C_m \times C_n \cong C_{mn}\cong \mathbb Z_{mn}$, a cyclic group.

So as you note, $C_{22}\times C_{33} \not\cong C_{22\cdot 33}$ and is not cyclic precisely because $\gcd(22, 33) = 11 \neq 1$.


As Jared very aptly commented: "the product of finitely many cyclic groups $C_{n_1}\times C_{n_2}\times\ldots\times C_{n_m}$ is cyclic if and only if the $n_i$ are pairwise relatively prime": if and only if $\gcd(n_i, n_j) = 1$ for all $i, j,\;\; 1\leq i,j, m,\;\;i\neq j$. And since the orders of such groups would be pairwise relatively prime, the order of the the group which is the direct product is $\;n_1\cdot n_2\cdot \ldots \cdot n_m$

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Hi Mark, here is an extra observation that you might find helpful following from amWhy's last statement: the product of finitely many cyclic groups $C_{n_1}\times C_{n_2}\times\ldots\times C_{n_m}$ is cyclic if and only if the $n_i$ are pairwise relatively prime. –  Jared Mar 18 '13 at 1:14
    
Thanks for generalizing, @Jared ! –  amWhy Mar 18 '13 at 1:57

If gcd$(m, n)=1$, we can say that $C_m\times C_n\cong C_{mn}$. The proof is simple, there must be an element $a\in C_m$ of order $m$ and an element $b\in C_n$ of order $n$. Then, $(a,b)$ will have order $mn$, which means $C_m\times C_n$ must be cyclic.

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Where did you use the assumption that gcd(m,n) = 1? –  Mark Mar 18 '13 at 1:18
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@Mark only when $(m,n)=1$, $(a,b)$ have order $mn$. Otherwise, the order of $(a, b)$ will be the least common multiple of $m$ and $n$, which is not $mn$. –  NECing Mar 18 '13 at 1:19

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