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Let $\langle \Bbb Z_{n \cdot m};+,0 \rangle$ be the additive group of integers modulo $m \cdot n$ and $\Bbb Z_{m} \times \Bbb Z_{n}$ the direct product of the two additive group of integers modulo $m$ and $n$ respectively.

Now I know that the following holds because of the CRT: $$ gcd(m, n) = 1 \implies \Bbb Z_{n \cdot m} \cong \Bbb Z_{m} \times \Bbb Z_{n} $$

What I want to know now if the other direction can also be proven (or disproven): $$ gcd(m, n) = 1 \impliedby \Bbb Z_{n \cdot m} \cong \Bbb Z_{m} \times \Bbb Z_{n} $$

So that if you have the direct product of two groups $\Bbb Z_{m}$ and $\Bbb Z_{n}$, it is only isomorphic to $\Bbb Z_{n \cdot m}$ if $n$ and $m$ are relatively prime.

I am looking for a proof for the second implication or a counter example if the second implication does not hold.

If this is provable, does the more general form

$$ \forall 1 \le i \le n , 1 \le j \le n: i \neq j \to gcd(m_{i},m_{j}) = 1 $$ $$ \iff $$ $$ \Bbb Z_{m_{1}} \times Z_{m_{2}} \times \dotsb \times Z_{m_{n}} \cong \Bbb Z_{m_{1} \cdot m_{2} \dotsb m_{n}} $$

also hold?

I strongly suspect that the implication holds in both directions but I have not found a proof for or a counter example against it.

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$\newcommand{\lcm}{\operatorname{lcm}}$ Suppose $\Bbb{Z}_{n\cdot m} \cong \Bbb{Z}_m\times \Bbb{Z}_n$, then the highest order of an element of $\Bbb{Z}_{n\cdot m}$ is $nm$, and the highest order of an element of $\Bbb{Z}_m\times \Bbb{Z}_n$ is $\lcm(n,m)$. Thus for the two groups to be isomorphic, we must have $\lcm(n,m)=nm$, but since $g\ell = nm$ where $g=\gcd(n,m)$, and $\ell = \lcm(n,m)$, this means that $g=1$, and so we must have had that $n$ and $m$ are relatively prime.

The larger case should be true by induction in one direction and a direct generalization of this argument in the other.

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    To prove the generalized case from this, it seems one can use the very same argumentation but just with the gcd and the lcm for $n$ integers instead of two.2017-01-04
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    Well you use the lcm for $n$ integers, but not the gcd for $n$ integers. The statement that 3 integers together have gcd 1 is not the same as being pairwise coprime: take 6, 10, 15. But if any two of them have a nontrivial common factor then the lcm will not be the product, so the argument still holds.2017-01-04
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    Yes I goofed there a little. If I see this correctly the product of all the $m_{i}$s is the lcm of all of them multiplied by the product of all the gcds of all the different pairs.2017-01-04
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    I mean I'm not sure if it is or not, but we know that when taking the LCM we can replace a subset of the numbers we are taking the LCM of with their LCM and get the same result, so in order for the total LCM to be the product, we need every pairwise LCM to be the product, or every pair of numbers to be coprime.2017-01-04
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    Yeah that makes sense, thanks :)2017-01-04