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In an exercise I did,I proved $$\mathbb{Z}_{m}\times\mathbb{Z}_{n}\cong\mathbb{Z}_{lcd(m,n)}\times\mathbb{Z}_{gcd(m,n)}.$$ I also like to show this kind of decomposition is unique,that's to say if $r|s$ and $\mathbb{Z}_{m}\times\mathbb{Z}_{n}\cong\mathbb{Z}_{r}\times\mathbb{Z}_{s}$ then $r=gcd(m,n)$ and $s=lcd(m,b).$

For example,how can I prove $\mathbb{Z}_{2}\times\mathbb{Z}_{16}$ is not isomorphic to $\mathbb{Z}_{4}\times\mathbb{Z}_{8}$?I think a simple case will be a hint.

Any help will be thanked.

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Hint: $\mathbb Z_2\times \mathbb Z_{16}$ has an element of order $16$. Does $\mathbb Z_4\times\mathbb Z_8$ also have one?

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To answer the question in the body:

If $r\mid s$ and $\mathbb{Z}_{m}\times\mathbb{Z}_{n}\cong\mathbb{Z}_{r}\times\mathbb{Z}_{s}$, then $r=gcd(m,n)$ and $s=lcd(m,b).$

Suppose $G=\mathbb{Z}_{m}\times\mathbb{Z}_{n}\cong\mathbb{Z}_{r}\times\mathbb{Z}_{s}$, with $r \mid s$.

Then $s=\exp(G)=lcm(m,n)$ and so $r = \dfrac{|G|}{s} = \dfrac{mn}{lcm(m,n)}=gcd(m,n)$.

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