# If $\gcd(m,n)=1$, then $\mathbb{Z}_n \times \mathbb{Z}_m$ is cyclic. [duplicate]

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If $\gcd(m,n)=1$, then $\mathbb{Z}_n \times \mathbb{Z}_m$ is a cyclic group.

Let's denote $\mathbb{Z}_n=\langle1_n \rangle$ and $\mathbb{Z}_m=\langle1_m \rangle.$ My proof goes as follows: since $|\mathbb{Z}_n \times \mathbb{Z}_m|=mn$ and $\gcd(m,n)=1$, $|\langle 1_n,1_m\rangle|=\text{lcm}(|\langle 1_n\rangle|,|\langle 1_m\rangle|)=mn$. Hence the direct product is cyclic. Is my proof correct?

## marked as duplicate by azimut, Ayman Hourieh, user1729, Aang, Nick PetersonJul 18 '13 at 10:28

• sorry yeah this is correct dont know what i was thinking – Ben Mar 2 '13 at 13:03
• I don't think this is a duplicate. The OP is not asking for a proof of the proposition, but whether his/her proof is correct or not. – user1551 Jul 18 '13 at 10:24

Your proof is correct if you actually know the key step

$$|(1_m, 1_n)| = \text{lcm}(|1_m|, |1_n|)$$

If one of my colleagues stated this I would believe they know what they're doing, but from a student I might be skeptical. You should probably explain why you think this is true, unless you think that should be common knowledge.

(Depending on how you justify it, you may have wanted $\geq$ there instead of $=$)

Anyways, this statement is a special case of the Chinese Remainder Theorem.

Yes, it works.

In general, if $R$ is a commutative ring with $1_R$ and $I$, $J$ are coprime ideals (i.e. $I+J = R$) then the map $R \to R\,/\,I \oplus R\,/\,J$ is surjective (Chinese Remainder Theorem).

As a consequence the ring $R\,/\,I \oplus R\,/\,J$ is cyclic.

• I mean cyclic as an $R$-module over itself, i.e. generated by one element. – Ivo Mar 2 '13 at 13:16