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Prove that $\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}$ implies $\gcd(m,n)=1$.

This is the converse of the Chinese remainder theorem in abstract algebra. Any help would be appreciated.

Thanks!

uh1
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    if $(m,n)=d\neq1$ then $C_n\times C_m$ has two subgroups of order $d$, whereas $C_{mn}$ only has one (looking at the additive structure if you're talking about rings) – yoyo Apr 28 '13 at 16:24
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    @DonAntonio : how is that a duplicate? Yes, it deals with very similar material but it certainly wasn't the same question... – Coffee_Table Apr 28 '13 at 16:38
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    @DonAntonio Note that your proposed duplicate is a converse of this question. But I think there are prior questions on this direction. – Math Gems Apr 28 '13 at 16:48

3 Answers3

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Suppose $\gcd(m, n) = d\gt 1$.

Then $\frac{mn}{d}$ is divisible by both $\,m \text { and}\;n.\;$ Therefore, for any $(r, s) \in \mathbb Z_m \times \mathbb Z_n$ we have that $$\underbrace{(r, s) + (r, s) + \cdots + (r, s)}_{\Large \frac{mn}{d} \,\lt\, mn \;\;\text{summands}} = (0, 0).$$ Hence no element $\,(r, s) \in \mathbb Z_m \times \mathbb Z_n\,$ can generate the entire group, whose order is $\,mn > \frac{mn}{d}.\,$ So, $\,\mathbb Z_m \times \mathbb Z_n\,$ is not cyclic, and thus, cannot be isomorphic to $\mathbb Z_{mn}$, a group we know is cyclic.

This proves the contrapositive of the claim. It proves : $$ \gcd(m, n) \neq 1 \implies \mathbb Z_m\times \mathbb Z_n \not\cong \mathbb Z_{mn}$$

Since an implication and its contrapositive are logically equivalent, we have thus shown:

$$\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}\,\text{ implies } \;\;\gcd(m,n)=1.$$

amWhy
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  • but even if $d=1$ $mn/d$ is still divisible by both $m,n.$ – uh1 Apr 28 '13 at 16:45
  • Yes, but if $d = 1, mn/d = mn.$ And $mn$ summands means the group $\mathbb Z_m \times \mathbb Z_n$ is the group cyclic and hence isomorphic to $\mathbb Z_{mn}$. Only when $d>1$ is $mn/d \lt mn$, hence $\mathbb Z_m \times \mathbb Z_n$ is not cyclic, and so not isomorphic to $\mathbb Z_{mn}$ – amWhy Apr 28 '13 at 16:48
  • Are all $mathbb Z_{n}$ groups cyclic? – uh1 Apr 28 '13 at 17:26
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    Yes, all groups of the form $\mathbb Z_n$ are cyclic. But not all groups of the form $\mathbb Z_m \times \mathbb Z_n$ are cyclic. – amWhy Apr 28 '13 at 17:27
  • @amWhy: Excellent guidance and nice to get acked! +1 You've been busy! ;-) – Amzoti Apr 29 '13 at 00:35
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Approach: Contrapositively, $(m,n)>1 \implies \mathbb{Z}_m\times\mathbb{Z}_n$ not cyclic $\implies \mathbb{Z}_m\times\mathbb{Z}_n \ncong \mathbb{Z}_{nm}$ since the latter is cyclic.

Basically, the contrapositive is much easier to verify (which is fine for our purposes, since it is equivalent); all one needs to do is to find some "basic property", i.e., one that should hold under isomorphism, and show that these two groups do not share the property. The one I chose is "being cyclic". (I leave it to you justify the steps I gave though.)

Coffee_Table
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Hint: Every element of $\mathbb{Z}_m\times \mathbb{Z}_n$ has order $\le$ the lcm of $m$ and $n$.

André Nicolas
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